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Hypothesis testing

Null and alternative hypotheses, test statistics, p-values, and the two kinds of error.

A hypothesis test asks whether the data are surprising if the null hypothesis were true. Compute how many standard errors the sample lies from the null value (the test statistic), turn that into a p-value, and reject the null when the p-value is below the significance level. A Type I error rejects a true null; a Type II error keeps a false one.

Null and Alternative Hypotheses

The actual test begins by considering two hypotheses. They are called the null hypothesis and the alternative hypothesis. These hypotheses contain opposing viewpoints.

H₀, the null hypothesis: a statement of no difference between sample means or proportions or no difference between a sample mean or proportion and a population mean or proportion. In other words, the difference equals 0.

H_a, the alternative hypothesis: a claim about the population that is contradictory to H₀ and what we conclude when we reject H₀.

Since the null and alternative hypotheses are contradictory, you must examine evidence to decide if you have enough evidence to reject the null hypothesis or not. The evidence is in the form of sample data.

After you have determined which hypothesis the sample supports, you make a decision. There are two options for a decision. They are reject H₀ if the sample information favors the alternative hypothesis or do not reject H₀ or decline to reject H₀ if the sample information is insufficient to reject the null hypothesis.

Mathematical Symbols Used in H₀ and H_a:

H₀	H_a
equal (=)	not equal (≠) or greater than (>) or less than (<)
greater than or equal to (≥)	less than (<)
less than or equal to (≤)	more than (>)

Example

H₀: No more than 30 percent of the registered voters in Santa Clara County voted in the primary election. p ≤ 30
H_a: More than 30 percent of the registered voters in Santa Clara County voted in the primary election. p > 30

Example

We want to test whether the mean GPA of students in American colleges is different from 2.0 (out of 4.0). The null and alternative hypotheses are the following:
H₀: μ = 2.0
H_a: μ ≠ 2.0

Example

We want to test if college students take fewer than five years to graduate from college, on the average. The null and alternative hypotheses are the following:
H₀: μ ≥ 5
H_a: μ < 5

Condensed — the full section is in OpenStax Statistics.

Outcomes and the Type I and Type II Errors

When you perform a hypothesis test, there are four possible outcomes depending on the actual truth, or falseness, of the null hypothesis H₀ and the decision to reject or not. The outcomes are summarized in the following table:

ACTION	H₀ IS ACTUALLY	...
	True	False
Do not reject H₀	Correct outcome	Type II error
Reject H₀	Type I error	Correct outcome

The four possible outcomes in the table are as follows:

The decision is not to reject H₀ when H₀ is true (correct decision).
The decision is to reject H₀ when, in fact, H₀ is true (incorrect decision known as a Type I error).
The decision is not to reject H₀ when, in fact, H₀ is false (incorrect decision known as a Type II error).
The decision is to reject H₀ when H₀ is false (correct decision whose probability is called the Power of the Test).

Each of the errors occurs with a particular probability. The Greek letters α and β represent the probabilities.

α = probability of a Type I error = P(Type I error) = probability of rejecting the null hypothesis when the null hypothesis is true.

β = probability of a Type II error = P(Type II error) = probability of not rejecting the null hypothesis when the null hypothesis is false.

α and β should be as small as possible because they are probabilities of errors. They are rarely zero.

The Power of the Test is 1 – β. Ideally, we want a high power that is as close to one as possible. Increasing the sample size can increase the Power of the Test.

Example

Suppose the null hypothesis, H₀, is: Frank's rock climbing equipment is safe.

Type I error: Frank does not go rock climbing because he considers that the equipment is not safe, when in fact, the equipment is really safe. Frank is making the mistake of rejecting the null hypothesis, when the equipment is actually safe!

Type II error: Frank goes climbing, thinking that his equipment is safe, but this is a mistake, and he painfully realizes that his equipment is not as safe as it should have been. Frank assumed that the null hypothesis was true, when it was not.

α = probability that Frank thinks his rock climbing equipment may not be safe when, in fact, it really is safe. β = probability that Frank thinks his rock climbing equipment may be safe when, in fact, it is not safe.

Notice that, in this case, the error with the greater consequence is the Type II error. (If Frank thinks his rock climbing equipment is safe, he will go ahead and use it.)

Condensed — the full section is in OpenStax Statistics.

Distribution Needed for Hypothesis Testing

Earlier in the course, we discussed sampling distributions. Particular distributions are associated with hypothesis testing. Perform tests of a population mean using a normal distribution or a Student's t-distribution. (Remember, use a Student's t-distribution when the population standard deviation is unknown and the distribution of the sample mean is approximately normal.) We perform tests of a population proportion using a normal distribution (usually n is large).

Assumptions

When you perform a hypothesis test of a single population mean μ using a Student's t-distribution (often called a t-test), there are fundamental assumptions that need to be met in order for the test to work properly. Your data should be a simple random sample that comes from a population that is approximately normally distributed. You use the sample standard deviation to approximate the population standard deviation. Note that if the sample size is sufficiently large, a t-test will work even if the population is not approximately normally distributed.

When you perform a hypothesis test of a single population mean μ using a normal distribution (often called a z-test), you take a simple random sample from the population. The population you are testing is normally distributed or your sample size is sufficiently large. You know the value of the population standard deviation which, in reality, is rarely known.

When you perform a hypothesis test of a single population proportion p, you take a simple random sample from the population. You must meet the conditions for a binomial distribution, which are the following: there are a certain number n of independent trials, the outcomes of any trial are success or failure, and each trial has the same probability of a success p. The shape of the binomial distribution needs to be similar to the shape of the normal distribution. To ensure this, the quantities np and nq must both be greater than five (np > 5 and nq > 5). Then the binomial distribution of a sample (estimated) proportion can be approximated by the normal distribution with μ = p and $\sigma =\sqrt{\frac{pq}{n}}$. Remember that q = 1 – p.

Rare Events, the Sample, and the Decision and Conclusion

Establishing the type of distribution, sample size, and known or unknown standard deviation can help you figure out how to go about a hypothesis test. However, there are several other factors you should consider when working out a hypothesis test.

Rare Events

The thinking process in hypothesis testing can be summarized as follows: You want to test whether or not a particular property of the population is true. You make an assumption about the true population mean for numerical data or the true population proportion for categorical data. This assumption is the null hypothesis. Then you gather sample data that is representative of the population. From this sample data you compute the sample mean (or the sample proportion). If the value that you observe is very unlikely to occur (a rare event) if the null hypothesis is true, then you wonder why this is happening. A plausible explanation is that the null hypothesis is false.

For example, Didi and Ali are at a birthday party of a very wealthy friend. They hurry to be first in line to grab a prize from a tall basket that they cannot see inside because they will be blindfolded. There are 200 plastic bubbles in the basket, and Didi and Ali have been told that there is only one with a $100 bill. Didi is the first person to reach into the basket and pull out a bubble. Her bubble contains a $100 bill. The probability of this happening is $\frac{1}{200}$ = 0.005. Because this is so unlikely, Ali is hoping that what the two of them were told is wrong and there are more $100 bills in the basket. A rare event has occurred (Didi getting the $100 bill) so Ali doubts the assumption about only one $100 bill being in the basket.

Using the Sample to Test the Null Hypothesis

After you collect data and obtain the test statistic (the sample mean, sample proportion, or other test statistic), you can determine the probability of obtaining that test statistic when the null hypothesis is true. This probability is called the p-value.

When the p-value is very small, it means that the observed test statistic is very unlikely to happen if the null hypothesis is true. This gives significant evidence to suggest that the null hypothesis is false, and to reject it in favor of the alternative hypothesis. In practice, to reject the null hypothesis we want the p-value to be smaller than 0.05 (5 percent) or sometimes even smaller than 0.01 (1 percent).

Condensed — the full section is in OpenStax Statistics.

Decision and Conclusion

A systematic way to make a decision of whether to reject or not reject the null hypothesis is to compare the p-value and a preset or preconceived α, also called the level of significance of the test. A preset α is the probability of a Type I error (rejecting the null hypothesis when the null hypothesis is true). It may or may not be given to you at the beginning of the problem.

When you make a decision to reject or not reject H₀, do as follows:

If p-value $<\alpha$, reject H₀. The results of the sample data are significant. There is sufficient evidence to conclude that H₀ is an incorrect belief and that the alternative hypothesis, H_a, may be correct.
If p-value $\ge \alpha$, do not reject H₀. The results of the sample data are not significant.There is not sufficient evidence to conclude that the alternative hypothesis, H_a, may be correct.
When you do not reject H₀, it does not mean that you should believe that H₀ is true. It simply means that the sample data have failed to provide sufficient evidence to cast serious doubt about the truthfulness of H₀.

Conclusion: After you make your decision, write a thoughtful conclusion about the hypotheses in terms of the given problem.

Example

When using the p-value to evaluate a hypothesis test, you might find it useful to use the following mnemonic device:

If the p-value is low, the null must go.

If the p-value is high, the null must fly.

This memory aid relates a p-value less than the established alpha (the p is low) as rejecting the null hypothesis and, likewise, relates a p-value higher than the established alpha (the p is high) as not rejecting the null hypothesis.

Try it.

Fill in the blanks.

Reject the null hypothesis when ______________________________________.

The results of the sample data _____________________________________.

Do not reject the null hypothesis when __________________________________________.

The results of the sample data ____________________________________________.

Solution

Reject the null hypothesis when the p-value is less than the established alpha value. The results of the sample data support the alternative hypothesis.

Do not reject the null hypothesis when the p-value is greater or equal to the established alpha value. The results of the sample data do not support the alternative hypothesis.

Condensed — the full section is in OpenStax Statistics.

Rare Events, the Sample, and the Decision and Conclusion

Use the following information to answer the next seven exercises: Suppose that a recent article stated that the mean time students spend doing homework each week is 2.5 hours. A study was then done to see if the mean time has increased in the new century. A random sample of 26 students was taken. The mean length of time they did homework each week was three hours with a standard deviation of 1.8 hours. Suppose that it is somehow known that the population standard deviation is 1.5. Conduct a hypothesis test to determine if the mean length of time doing homework each week has increased. Assume the distribution of homework times is approximately normal.

Additional Information and Full Hypothesis Test Examples

In a hypothesis test problem, you may see words such as "the level of significance is 1 percent". The "1 percent" is the preconceived or preset α.
The statistician setting up the hypothesis test selects the value of α to use before collecting the sample data.
If no level of significance is given, a common standard to use is α = 0.05.
When you calculate the p-value and draw the picture, the p-value is the area in the left tail, the right tail, or split evenly between the two tails. For this reason, we call the hypothesis test left, right, or two tailed.
The alternative hypothesis, ${H}_{a}$, tells you if the test is left, right, or two-tailed. It is the key to conducting the appropriate test.
H_a never has a symbol that contains an equal sign.
Thinking about the meaning of the p-value: A data analyst should have more confidence that he made the correct decision to reject the null hypothesis with a smaller p-value (for example, 0.001 as opposed to 0.04) even if using the 0.05 level for alpha. Similarly, for a large p-value such as 0.4, as opposed to a p-value of 0.056 (alpha = 0.05 is less than either number), a data analyst should have more confidence that she made the correct decision in not rejecting the null hypothesis. This makes the data analyst use judgment rather than mindlessly applying rules.

The following examples illustrate a left-, right-, and two-tailed test.

Example

H₀: μ = 5 H_a: μ < 5

Test of a single population mean. H_a tells you the test is left-tailed. The picture of the p-value is as follows:

Example

H₀: p ≤ 0.2 H_a: p > 0.2

This is a test of a single population proportion. H_a tells you the test is right-tailed. The picture of the p-value is as follows:

Example

H₀: p = 50 H_a: p ≠ 50

This is a test of a single population mean. H_a tells you the test is two-tailed. The picture of the p-value is as follows.

Full Hypothesis Test Examples

Example

Try it.

Jeffrey, as an eight-year-old, established a mean time of 16.43 seconds for swimming the 25-yard freestyle, with a standard deviation of 0.8 seconds. His dad, Frank, thought that Jeffrey could swim the 25-yard freestyle faster using goggles. Frank bought Jeffrey a new pair of expensive goggles and timed Jeffrey for 15 25-yard freestyle swims. For the 15 swims, Jeffrey's mean time was 16 seconds. Frank thought that the goggles helped Jeffrey to swim faster than the 16.43 seconds. Conduct a hypothesis test using a preset α = 0.05. Assume that the swim times for the 25-yard freestyle are normal.

Solution

Set up the hypothesis test:

Since the problem is about a mean, this is a test of a single population mean.

H₀: μ = 16.43 H_a: μ < 16.43

For Jeffrey to swim faster, his time will be less than 16.43 seconds. The "<" tells you this is left-tailed.

Determine the distribution needed:

Random variable: $X$ = the mean time to swim the 25-yard freestyle.

Distribution for the test: $X$ is normal (population standard deviation is known: σ = 0.8)

with mean $\mu =16.43$ and standard error of $\frac{0.8}{\sqrt{15}}$;

μ = 16.43 comes from H₀ and not the data. σ = 0.8, and n = 15.

Using a table or a calculator, we can calculate the p-value as the area to the left of 16 under the normal curve:

p-value = P($\overset{\bar}{x}$ < 16) = 0.0187 where the sample mean in the problem is given as 16.

p-value = 0.0187. The p-value is the area to the left of the sample mean given as 16.

Graph:

μ = 16.43 comes from H₀. Our assumption is μ = 16.43.

Interpretation of the p-value: If H₀ is true, there is a 0.0187 probability (1.87 percent), that Jeffrey's mean time to swim the 25-yard freestyle is 16 seconds or less. Because a 1.87 percent chance is small, the mean time of 16 seconds or less is unlikely to have happened randomly. It is a rare event.

Compare α and the p-value:

α = 0.05 p-value = 0.0187 α > p-value

Make a decision: Since α > p-value, reject H₀.

An alternative approach is to find the z-test corresponding to the sample mean $\overset{\bar}{x}=16.$ This is \[z\text{-test}=\frac{\overset{\bar}{x}-{\mu }_{X}}{\frac{{\sigma }_{X}}{\sqrt{n}}}=\frac{16-16.43}{\frac{0.8}{\sqrt{15}}}=-2.081729\].

The critical z-value = –1.645 for this test has probability 0.05 to its left tail, according to the Normal Table (see Appendices). Because the z-test is to the left of the critical z-value, we reject the null hypothesis.

This means that you reject μ = 16.43. In other words, you do not think Jeffrey swims the 25-yard freestyle in 16.43 seconds but instead that he swims faster with the new goggles.

Conclusion: At the 5 percent significance level, we conclude that Jeffrey swims faster using the new goggles. The sample data show there is sufficient evidence that Jeffrey's mean time to swim the 25-yard freestyle is less than 16.43 seconds.

The p-value can easily be calculated.

When the calculator does a z-Test, the z-Test function finds the p-value by doing a normal probability calculation:

$P(\overset{\bar}{x}<16)=$ 2nd DISTR normcdf $(-10\wedge 99,16,16.43,0.8/\sqrt{15})\text{.}$

The Type I and Type II errors for this problem are as follows:

The Type I error is to conclude that Jeffrey swims the 25-yard freestyle, on average, in less than 16.43 seconds when, in fact, he actually swims the 25-yard freestyle, on average, in 16.43 seconds. (Reject the null hypothesis when the null hypothesis is true.)

The Type II error is that there is not evidence to conclude that Jeffrey swims the 25-yard freestyle, on average, in less than 16.43 seconds when, in fact, he actually does swim the 25-yard freestyle, on average, in less than 16.43 seconds. (Do not reject the null hypothesis when the null hypothesis is false.)

Condensed — the full section is in OpenStax Statistics.

လုပ်ဆောင်ခဲ့သောဥပမာ: `(52 - 50)/(10/sqrt(25))`

Arithmetic · Arithmetic

Evaluate 1

ခြေလှမ်းတစ်လှမ်း

\frac{52 - 50}{10 \frac{1}{\sqrt{25}}} = 1
Add: 52 - 50 = 52 - 50.

အဖြေကို ဖော်ပြပါ

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

A medical trial is conducted to test whether or not a new medicine reduces cholesterol by 25 percent. State the null and alternative hypotheses.

အဖြေကို ဖော်ပြပါ

H₀ : The drug reduces cholesterol by 25 percent. p = 0.25
H_a : The drug does not reduce cholesterol by 25 percent. p ≠ 0.25
We want to test whether the mean height of eighth graders is 66 inches. State the null and alternative hypotheses. Fill in the correct symbol (=, ≠, ≥, <, ≤, >) for the null and alternative hypotheses.
1. H₀: μ __ 66
2. H_a: μ __ 66
အဖြေကို ဖော်ပြပါ
1. H₀ : μ = 66
2. H_a : μ ≠ 66
We want to test if it takes fewer than 45 minutes to teach a lesson plan. State the null and alternative hypotheses. Fill in the correct symbol ( =, ≠, ≥, <, ≤, >) for the null and alternative hypotheses.
1. H₀: μ __ 45
2. H_a: μ __ 45
အဖြေကို ဖော်ပြပါ
1. H₀: μ ≥ 45
2. H_a: μ < 45
On a state driver’s test, about 40 percent pass the test on the first try. We want to test if more than 40 percent pass on the first try. Fill in the correct symbol (=, ≠, ≥, <, ≤, >) for the null and alternative hypotheses.
1. H₀: p __ 0.40
2. H_a: p __ 0.40
အဖြေကို ဖော်ပြပါ
1. H₀: p = 0.40
2. H_a: p > 0.40
You are testing that the mean speed of your cable internet connection is more than three megabits per second. What is the random variable? Describe it in words.

အဖြေကို ဖော်ပြပါ

The random variable is the mean Internet speed in megabits per second.
You are testing that the mean speed of your cable internet connection is more than three megabits per second. State the null and alternative hypotheses.
The American family has an average of two children. What is the random variable? Describe in words.

အဖြေကို ဖော်ပြပါ

The random variable is the mean number of children an American family has.
The mean entry level salary of an employee at a company is $58,000. You believe it is higher for IT professionals in the company. State the null and alternative hypotheses.
A sociologist claims the probability that a person picked at random in Times Square in New York City is visiting the area is 0.83. You want to test to see if the proportion is actually less. What is the random variable? Describe in words.

အဖြေကို ဖော်ပြပါ

The random variable is the proportion of people picked at random in Times Square visiting the city.
A sociologist claims the probability that a person picked at random in Times Square in New York City is visiting the area is 0.83. You want to test to see if the claim is correct. State the null and alternative hypotheses.
In a population of fish, approximately 42 percent are female. A test is conducted to see if, in fact, the proportion is less. State the null and alternative hypotheses.
အဖြေကို ဖော်ပြပါ
1. H₀: p = 0.42
2. H_a: p < 0.42
Suppose that a recent article stated that the mean time students spend doing homework each week is 2.5 hours. A study was then done to see if the mean time has increased in the new century. A random sample of 26 students. The mean length of time the students spent on homework was 3 hours with a standard deviation of 1.8 hours. Suppose that it is somehow known that the population standard deviation is 1.5. If you were conducting a hypothesis test to determine if the mean length of homework has increased, what would the null and alternative hypotheses be? The distribution of the population is normal.
1. H₀: ________
2. H_a: ________
A random survey of 75 long-term marathon runners revealed that the mean length of time they've been running is 17.4 years with a standard deviation of 6.3 years. If you were conducting a hypothesis test to determine if the population mean time for these runners could likely be 15 years, what would the null and alternative hypotheses be?
1. H₀: __________
2. H_a: __________
အဖြေကို ဖော်ပြပါ
1. H₀: μ = 15
2. H_a: μ ≠ 15
Researchers published an article stating that in any one-year period, approximately 9.5 percent of American adults suffer from a particular type of disease. Suppose that in a survey of 100 people in a certain town, seven of them suffered from this disease. If you were conducting a hypothesis test to determine if the true proportion of people in that town suffering from this disease is lower than the percentage in the general adult American population, what would the null and alternative hypotheses be?
1. H₀: ________
2. H_a: ________
Some of the following statements refer to the null hypothesis, some to the alternate hypothesis.
State the null hypothesis, H₀, and the alternative hypothesis. H_a, in terms of the appropriate parameter (μ or p).
1. The mean number of years Americans work before retiring is 34.
2. At most 60 percent of Americans vote in presidential elections.
3. The mean starting salary for San Jose State University graduates is at least $100,000 per year.
4. Twenty-nine percent of high school students take physical education daily.
5. Less than 5 percent of adults ride the bus to work in Los Angeles.
6. The mean number of cars a person owns in her lifetime is not more than 10.
7. About half of Americans prefer to live away from cities, given the choice.
8. Europeans have a mean paid vacation each year of six weeks.
9. The chance of developing breast cancer is under 11 percent for women.
10. Private universities' mean tuition cost is more than $20,000 per year.
အဖြေကို ဖော်ပြပါ
1. H₀: μ = 34; H_a: μ ≠ 34
2. H₀: p ≤ 0.60; H_a: p > 0.60
3. H₀: μ ≥ 100,000; H_a: μ < 100,000
4. H₀: p = 0.29; H_a: p ≠ 0.29
5. H₀: p = 0.05; H_a: p < 0.05
6. H₀: μ ≤ 10; H_a: μ > 10
7. H₀: p = 0.50; H_a: p ≠ 0.50
8. H₀: μ = 6; H_a: μ ≠ 6
9. H₀: p ≥ 0.11; H_a: p < 0.11
10. H₀: μ ≤ 20,000; H_a: μ > 20,000
A recent survey of 273 randomly selected teens living in Massachusetts asked about social media. Sixty-three said that they routinely use a certain app to share pictures. The researchers want to determine if there is good evidence that more than 30 percent of teens use this app. The alternative hypothesis is as follows:
1. p < 0.30
2. p ≤ 0.30
3. p ≥ 0.30
4. p > 0.30
A statistics instructor believes that fewer than 20 percent of Evergreen Valley College (EVC) students attended the opening night midnight showing of the latest Harry Potter movie. She surveys 84 of her students and finds that 11 attended the midnight showing. An appropriate alternative hypothesis is as follows:
1. p = 0.20
2. p > 0.20
3. p < 0.20
4. p ≤ 0.20
အဖြေကို ဖော်ပြပါ

c
Previously, an organization reported that teenagers spent 4.5 hours per week, on average, on the phone. The organization thinks that, currently, the mean is higher. Fifteen randomly chosen teenagers were asked how many hours per week they spend on the phone. The sample mean was 4.75 hours with a sample standard deviation of 2.0. Conduct a hypothesis test. The null and alternative hypotheses are as follows:
1. H_o: $\overset{\bar}{x}$ = 4.5, H_a : $\overset{\bar}{x}$ > 4.5
2. H_o: μ ≥ 4.5, H_a: μ < 4.5
3. H_o: μ = 4.75, H_a: μ > 4.75
4. H_o: μ = 4.5, H_a: μ > 4.5
Suppose the null hypothesis, H₀, is: the blood cultures contain no traces of pathogen X. State the Type I and Type II errors.

အဖြေကို ဖော်ပြပါ

Type I error: The researcher thinks the blood cultures do contain traces of pathogen X, when in fact, they do not.
Type II error: The researcher thinks the blood cultures do not contain traces of pathogen X, when in fact, they do.
Suppose the null hypothesis, H₀, is: a patient is not sick. Which type of error has the greater consequence, Type I or Type II?

အဖြေကို ဖော်ပြပါ

The error with the greater consequence is the Type II error: the patient will be thought well when, in fact, he is sick, so he will not get treatment.
Red tide is a bloom of poison-producing algae—a few different species of a class of plankton called dinoflagellates. When the weather and water conditions cause these blooms, shellfish such as clams living in the area develop dangerous levels of a paralysis-inducing toxin. In Massachusetts, the Division of Marine Fisheries montors levels of the toxin in shellfish by regular sampling of shellfish along the coastline. If the mean level of toxin in clams exceeds 800 μg (micrograms) of toxin per kilogram of clam meat in any area, clam harvesting is banned there until the bloom is over and levels of toxin in clams subside. Describe both a Type I and a Type II error in this context, and state which error has the greater consequence.

အဖြေကို ဖော်ပြပါ

In this scenario, an appropriate null hypothesis would be H₀: the mean level of toxins is at most 800 μg, H₀ : μ₀ ≤ 800 μg.
Type I error: The DMF believes that toxin levels are still too high when, in fact, toxin levels are at most 800 μg. The DMF continues the harvesting ban.
Type II error: The DMF believes that toxin levels are within acceptable levels (are at least 800 μg) when, in fact, toxin levels are still too high (more than 800 μg). The DMF lifts the harvesting ban. This error could be the most serious. If the ban is lifted and clams are still toxic, consumers could possibly eat tainted food.
In summary, the more dangerous error would be to commit a Type II error, because this error involves the availability of tainted clams for consumption.
Identify the Type I and Type II errors from these four possible choices.
1. Not to reject the null hypothesis that the percentage of adults who have jobs is at least 88 percent when that percentage is actually less than 88 percent
2. Not to reject the null hypothesis that the percentage of adults who have jobs is at least 88 percent when the percentage is actually at least 88 percent
3. Reject the null hypothesis that the percentage of adults who have jobs is at least 88 percent when the percentage is actually at least 88 percent
4. Reject the null hypothesis that the percentage of adults who have jobs is at least 88 percent when that percentage is actually less than 88 percent
အဖြေကို ဖော်ပြပါ

Type I error: c
Type I error: b
The mean price of mid-sized cars in a region is $32,000. A test is conducted to see if the claim is true. State the Type I and Type II errors in complete sentences.

အဖြေကို ဖော်ပြပါ

Type I: The mean price of mid-sized cars is $32,000, but we conclude that it is not $32,000.
Type II: The mean price of mid-sized cars is not $32,000, but we conclude that it is $32,000.
A sleeping bag is tested to withstand temperatures of –15 °F. You think the bag cannot stand temperatures that low. State the Type I and Type II errors in complete sentences.
For Exercise 9.12, what are α and β in words?

အဖြေကို ဖော်ပြပါ

α = the probability that you think the bag cannot withstand 15 degrees F, when, in fact, it can.
β = the probability that you think the bag can withstand 15 degrees F, when, in fact, it cannot.
In words, describe 1 – β for Exercise 9.12.
A group of doctors is deciding whether or not to perform an operation. Suppose the null hypothesis, H₀, is: the surgical procedure will go well. State the Type I and Type II errors in complete sentences.

အဖြေကို ဖော်ပြပါ

Type I: The procedure will go well, but the doctors think it will not.
Type II: The procedure will not go well, but the doctors think it will.
A group of doctors is deciding whether or not to perform an operation. Suppose the null hypothesis, H₀, is: the surgical procedure will go well. Which is the error with the greater consequence?
The power of a test is 0.981. What is the probability of a Type II error?

အဖြေကို ဖော်ပြပါ

0.019
A group of divers is exploring an old sunken ship. Suppose the null hypothesis, H₀, is the sunken ship does not contain buried treasure. State the Type I and Type II errors in complete sentences.
A microbiologist is testing a water sample for E. coli. Suppose the null hypothesis, H₀, is the sample does not contain E. coli. The probability that the sample does not contain E. coli, but the microbiologist thinks it does is 0.012. The probability that the sample does contain E. coli, but the microbiologist thinks it does not is 0.002. What is the power of this test?

အဖြေကို ဖော်ပြပါ

0.998
A microbiologist is testing a water sample for E. coli. Suppose the null hypothesis, H₀, is the sample contains E-coli. Which is the error with the greater consequence?
State the Type I and Type II errors in complete sentences given the following statements.
1. The mean number of years Americans work before retiring is 34.
2. At most 60 percent of Americans vote in presidential elections.
3. The mean starting salary for San Jose State University graduates is at least $100,000 per year.
4. 29 percent of high school students take physical education every day.
5. Less than 5 percent of adults ride the bus to work in Los Angeles.
6. The mean number of cars a person owns in his or her lifetime is not more than 10.
7. About half of Americans prefer to live away from cities, given the choice.
8. Europeans have a mean paid vacation each year of six weeks.
9. The chance of developing breast cancer is under 11 percent for women.
10. Private universitie' mean tuition cost is more than $20,000 per year.
အဖြေကို ဖော်ပြပါ
1. Type I error: We conclude that the mean is not 34 years, when it really is 34 years. Type II error: We conclude that the mean is 34 years, when in fact it really is not 34 years.
2. Type I error: We conclude that more than 60 percent of Americans vote in presidential elections, when the actual percentage is at most 60 percent.Type II error: We conclude that at most 60 percent of Americans vote in presidential elections when, in fact, more than 60 percent do.
3. Type I error: We conclude that the mean starting salary is less than $100,000, when it really is at least $100,000. Type II error: We conclude that the mean starting salary is at least $100,000 when, in fact, it is less than $100,000.
4. Type I error: We conclude that the proportion of high school seniors who take physical education daily is not 29%, when it really is 29%. Type II error: We conclude that the proportion of high school seniors who take physical education daily is 29% when, in fact, it is not 29%.
5. Type I error: We conclude that fewer than 5 percent of adults ride the bus to work in Los Angeles, when the percentage that do is really 29%. Type II error: We conclude that 29%. or more adults ride the bus to work in Los Angeles when, in fact, fewer that 29% do.
6. Type I error: We conclude that the mean number of cars a person owns in his or her lifetime is more than 10, when in reality it is not more than 10. Type II error: We conclude that the mean number of cars a person owns in his or her lifetime is not more than 10 when, in fact, it is more than 10.
7. Type I error: We conclude that the proportion of Americans who prefer to live away from cities is not about half, though the actual proportion is about half. Type II error: We conclude that the proportion of Americans who prefer to live away from cities is half when, in fact, it is not half.
8. Type I error: We conclude that the duration of paid vacations each year for Europeans is not six weeks, when in fact it is six weeks. Type II error: We conclude that the duration of paid vacations each year for Europeans is six weeks when, in fact, it is not.
9. Type I error: We conclude that the proportion is less than 11 percent, when it is really at least 11 percent. Type II error: We conclude that the proportion of women who develop breast cancer is at least 11 percent, when in fact it is less than 11 percent.
10. Type I error: We conclude that the average tuition cost at private universities is more than $20,000, though in reality it is at most $20,000. Type II error: We conclude that the average tuition cost at private universities is at most $20,000 when, in fact, it is more than $20,000.
For Statements A–J in , answer the following in complete sentences.
1. State a consequence of committing a Type I error.
2. State a consequence of committing a Type II error.
When a new drug is created, the pharmaceutical company must subject it to testing before receiving the necessary permission from the U.S. Food and Drug Administration (FDA) to market the drug. Suppose the null hypothesis is the drug is unsafe. What is the Type II error?
1. To conclude the drug is safe when, in fact, it is unsafe.
2. Not to conclude the drug is safe when, in fact, it is safe.
3. To conclude the drug is safe when, in fact, it is safe.
4. Not to conclude the drug is unsafe when, in fact, it is unsafe.
အဖြေကို ဖော်ပြပါ

b
A statistics instructor believes that fewer than 20 percent of Evergreen Valley College (EVC) students attended the opening midnight showing of the latest Harry Potter movie. She surveys 84 of her students and finds that 11 of them attended the midnight showing. The Type I error is to conclude that the percent of EVC students who attended is ________.
1. at least 20 percent, when, in fact, it is less than 20 percent.
2. 20 percent, when, in fact, it is 20 percent.
3. less than 20 percent, when, in fact, it is at least 20 percent.
4. less than 20 percent, when, in fact, it is less than 20 percent.
It is believed that Lake Tahoe Community College (LTCC) Intermediate Algebra students get less than seven hours of sleep per night, on average. A survey of 22 LTCC Intermediate Algebra students generated a mean of 7.24 hours with a standard deviation of 1.93 hours. At a level of significance of 5 percent, do LTCC Intermediate Algebra students get less than seven hours of sleep per night, on average?
The Type II error is not to reject that the mean number of hours of sleep LTCC students get per night is at least seven when, in fact, the mean number of hours
1. is more than seven hours.
2. is at most seven hours.
3. is at least seven hours.
4. is less than seven hours.
အဖြေကို ဖော်ပြပါ

d
Previously, an organization reported that teenagers spent 4.5 hours per week, on average, on the phone. The organization thinks that, currently, the mean is higher. Fifteen randomly chosen teenagers were asked how many hours per week they spend on the phone. The sample mean was 4.75 hours with a sample standard deviation of 2.0. Conduct a hypothesis test. The Type I error is
1. to conclude that the current mean hours per week is higher than 4.5, when, in fact, it is higher.
2. to conclude that the current mean hours per week is higher than 4.5, when, in fact, it is the same.
3. to conclude that the mean hours per week currently is 4.5, when, in fact, it is higher.
4. to conclude that the mean hours per week currently is no higher than 4.5, when, in fact, it is not higher.
Which two distributions can you use for hypothesis testing for this chapter?

အဖြေကို ဖော်ပြပါ

A normal distribution or a Student’s t-distribution
Which distribution do you use when the standard deviation is not known? Assume sample size is large.

သင့်ရဲ့ကိုယ်ပိုင်စမ်းသပ်

Parts of this page are adapted from OpenStax Statistics (CC BY 4.0). Condensed and re-explained here; errors are ours.

ပိုပြီး Statistics & Probability

Sampling and data Describing data with graphs Mean, median and mode Probability Counting: permutations and combinations Discrete random variables Continuous random variables The normal distribution The central limit theorem Confidence intervals Comparing two samples Chi-square tests Linear regression and correlation ANOVA and the F distribution

Null and Alternative Hypotheses

Outcomes and the Type I and Type II Errors

Distribution Needed for Hypothesis Testing

Assumptions

Rare Events, the Sample, and the Decision and Conclusion

Rare Events

Using the Sample to Test the Null Hypothesis

Decision and Conclusion

Rare Events, the Sample, and the Decision and Conclusion

Additional Information and Full Hypothesis Test Examples

Full Hypothesis Test Examples

လုပ်ဆောင်ခဲ့သောဥပမာ: (52 - 50)/(10/sqrt(25))

Evaluate 1

ခြေလှမ်းတစ်လှမ်း

Practice (40)

သင့်ရဲ့ကိုယ်ပိုင်စမ်းသပ်

ပိုပြီး Statistics & Probability

လုပ်ဆောင်ခဲ့သောဥပမာ: `(52 - 50)/(10/sqrt(25))`