Chi-square tests

Facts About the Chi-Square Distribution

The notation for the chi-square distribution is

\[\chi ∼{\chi }_{df}^{2}\]

where df = degrees of freedom, which depends on how chi-square is being used. If you want to practice calculating chi-square probabilities then use df = n – –1. The degrees of freedom for the three major uses are calculated differently.

For the χ² distribution, the population mean is μ = df, and the population standard deviation is \(\sigma =\sqrt{2(df)}\).

The random variable is shown as χ², but it may be any uppercase letter.

The random variable for a chi-square distribution with k degrees of freedom is the sum of k independent, squared standard normal variables is

χ² = (Z₁)² + (Z₂)² + ... + (Z_k)², where the following are true:

• The curve is nonsymmetrical and skewed to the right.
• There is a different chi-square curve for each df.
• The test statistic for any test is always greater than or equal to zero.
• When df > 90, the chi-square curve approximates the normal distribution. For X ~ \({\chi }_{1,000}^{2}\), the mean, μ = df = 1,000 and the standard deviation, σ = \(\sqrt{2(1,000)}\) = 44.7. Therefore, X ~ N(1,000, 44.7), approximately.
• The mean, μ, is located just to the right of the peak.

Goodness-of-Fit Test

In this type of hypothesis test, you determine whether the data fit a particular distribution. For example, you may suspect your unknown data fit a binomial distribution. You use a chi-square test, meaning the distribution for the hypothesis test is chi-square, to determine if there is a fit. The null and the alternative hypotheses for this test may be written in sentences or may be stated as equations or inequalities.

The test statistic for a goodness-of-fit test is:

\[\underset{k}{\Sigma }\frac{{(O-E)}^{2}}{E}\]

where

• O = observed values (data),
• E = expected values (from theory), and
• k = the number of different data cells or categories.

The observed values are the data values, and the expected values are the values you would expect to get if the null hypothesis were true. There are n terms of the form \(\frac{{(O-E)}^{2}}{E}\).

The number of degrees of freedom is df = (number of categories – 1).

The goodness-of-fit test is almost always right-tailed. If the observed values and the corresponding expected values are not close to each other, then the test statistic can get very large and will be way out in the right tail of the chi-square curve.

Condensed — the full section is in OpenStax Statistics.

Goodness-of-Fit Test

Determine the appropriate test to be used in the next three exercises.

Use the following information to answer the next five exercises. A teacher predicts the distribution of grades on the final exam. The predictions are shown in .

Grade	Proportion
A	0.25
B	0.30
C	0.35
D	0.10

The actual distribution for a class of 20 is in .

Grade	Frequency
A	7
B	7
C	5
D	1

Use the following information to answer the next nine exercises. The cumulative number of cases of a chronic disease reported for Santa Clara County is broken down by ethnicity as in .

Ethnicity	Number of Cases
White	2,229
Hispanic	1,157
Black/African American	457
Asian, Pacific Islander	232
	Total = 4,075

The percentage of each ethnic group in Santa Clara County is as in .

Ethnicity	% of Total County Population	Number Expected (round to two decimal places)
White	42.9%	1,748.18
Hispanic	26.7%
Black/African American	2.6%
Asian, Pacific Islander	27.8%
	Total = 100%

Test of Independence

Tests of independence involve using a contingency table of observed (data) values.

The test statistic for a test of independence is similar to that of a goodness-of-fit test

\[\underset{(i⋅j)}{\Sigma }\frac{{(O-E)}^{2}}{E}\]

where

• O = observed values,
• E = expected values,
• i = the number of rows in the table, and
• j = the number of columns in the table.

There are \(i⋅j\) terms of the form \(\frac{{(O-E)}^{2}}{E}\).

A test of independence determines whether two factors are independent. You first encountered the term independence in Probability Topics. As a review, consider the following example.

Condensed — the full section is in OpenStax Statistics.

Test of Independence

Determine the appropriate test to be used in the next three exercises.

Use the following information to answer the next seven exercises: Transit Railroads is interested in the relationship between travel distance and the ticket class purchased. A random sample of 200 passengers is taken. shows the results. The railroad wants to know if a passenger’s choice in ticket class is independent of the distance the passenger must travel.

Traveling Distance	Third Class	Second Class	First Class	Total
1–100 miles	21	14	6	41
101–200 miles	18	16	8	42
201–300 miles	16	17	15	48
301–400 miles	12	14	21	47
401–500 miles	6	6	10	22
Total	73	67	60	200

Use the following information to answer the next ten exercises. An article in the New England Journal of Medicine discussed a study on people who used a certain product in California and Hawaii. In one part of the report, the self-reported ethnicity and product-use levels per day were given. Of the people using the product at most 10 times per day, there were 9,886 African Americans, 2,745 Native Hawaiians, 12,831 Latinos, 8,378 Japanese Americans, and 7,650 whites. Of the people using the product 11 to 20 times per day, there were 6,514 African Americans, 3,062 Native Hawaiians, 4,932 Latinos, 10,680 Japanese Americans, and 9,877 whites. Of the people using the product 21 to 30 times per day, there were 1,671 African Americans, 1,419 Native Hawaiians, 1,406 Latinos, 4,715 Japanese Americans, and 6,062 whites. Of the people using the product at least 31 times per day, there were 759 African Americans, 788 Native Hawaiians, 800 Latinos, 2,305 Japanese Americans, and 3,970 whites.

State the decision and conclusion (in a complete sentence) for the following levels of α.

Test for Homogeneity

The goodness-of-fit test can be used to decide whether a population fits a given distribution, but it will not suffice to decide whether two populations follow the same unknown distribution. A different test, called the test for homogeneity, can be used to draw a conclusion about whether two populations have the same distribution. To calculate the test statistic for a test for homogeneity, follow the same procedure as with the test of independence.

H₀: The distributions of the two populations are the same.

H_a: The distributions of the two populations are not the same.

Use a \({\chi }^{2}\) test statistic. It is computed in the same way as the test for independence.

df = (number of columns – 1)(number of rows – 1)

All values in the table must be greater than or equal to five.

Comparing two populations. For example: men vs. women, before vs. after, east vs. west. The variable is categorical with more than two possible response values.

Condensed — the full section is in OpenStax Statistics.

Test for Homogeneity

Use the following information to answer the next five exercises. Do private practice doctors and hospital doctors have the same distribution of working hours? Suppose that a sample of 100 private practice doctors and 150 hospital doctors are selected at random and asked about the number of hours a week they work. The results are shown in .

	20–30	30–40	40–50	50–60
Private Practice	16	40	38	6
Hospital	8	44	59	39

Test of a Single Variance

A test of a single variance assumes that the underlying distribution is normal. The null and alternative hypotheses are stated in terms of the population variance or population standard deviation. The test statistic is

\[\frac{(n-1){s}^{2}}{{\sigma }^{2}}\]

where

• n = the total number of data,
• s² = sample variance, and
• σ² = population variance.

You may think of s as the random variable in this test. The number of degrees of freedom is df = n – 1. A test of a single variance may be right-tailed, left-tailed, or two-tailed. will show you how to set up the null and alternative hypotheses. The null and alternative hypotheses contain statements about the population variance.

Example

Try it.

Math instructors are not only interested in how their students do on exams, on average, but how the exam scores vary. To many instructors, the variance, or standard deviation, may be more important than the average.

Suppose a math instructor believes that the standard deviation for his final exam is five points. One of his best students thinks otherwise. The student claims that the standard deviation is more than five points. If the student were to conduct a hypothesis test, what would the null and alternative hypotheses be?

Solution

Even though we are given the population standard deviation, we can set up the test using the population variance as follows:

• H₀: σ² = 5²
• H_a: σ² > 5²

Condensed — the full section is in OpenStax Statistics.

Test of a Single Variance

Use the following information to answer the next three exercises. An archer’s standard deviation for his hits is six, where the data are measured in distance from the center of the target. An observer claims the standard deviation is less than six.

Use the following information to answer the next three exercises. The standard deviation of heights for students in a school is 0.81. A random sample of 50 students is taken, and the standard deviation of heights of the sample is 0.96. A researcher in charge of the study believes the standard deviation of heights for the school is greater than 0.81.

Use the following information to answer the next four exercises: The average waiting time in a doctor’s office varies. The standard deviation of waiting times in a doctor’s office is 3.4 minutes. A random sample of 30 patients in the doctor’s office has a standard deviation of waiting times of 4.1 minutes. One doctor believes the variance of waiting times is greater than originally thought.

કામ કરેલ ઉદાહરણ: (20-25)^2/25 + (30-25)^2/25

Practice (40)

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

1. If the number of degrees of freedom for a chi-square distribution is 25, what is the population mean and standard deviation?

   જવાબ બતાવો

   mean = 25 and standard deviation = 7.0711

2. If df > 90, the distribution is _____________. If df = 15, the distribution is ________________.

3. When does the chi-square curve approximate a normal distribution?

   જવાબ બતાવો

   when the number of degrees of freedom is greater than 90

4. Where is μ located on a chi-square curve?

5. Is it more likely the df is 90, 20, or 2 in the graph?

   જવાબ બતાવો

   df = 2

6. As the number of degrees of freedom increases, the graph of the chi-square distribution looks more and more symmetrical.

   જવાબ બતાવો

   true

7. The standard deviation of the chi-square distribution is twice the mean.

8. The mean and the median of the chi-square distribution are the same if df = 24.

   જવાબ બતાવો

   false

9. a. Can you use the information as it appears in the charts to conduct the goodness-of-fit test?

   જવાબ બતાવો

   a. No. Notice that the expected number of absences for the 12+ entry is less than five; it is two. Combine that group with the 9–11 group to create new tables where the number of students for each entry is at least five. The new results are in and .
   

   Number of Absences per Term	Expected Number of Students
   0–2	50
   3–5	30
   6–8	12
   9+	8

   Number of Absences per Term	Actual Number of Students
   0–2	35
   3–5	40
   6–8	20
   9+	5

10. b. What is the number of degrees of freedom (df)?

    જવાબ બતાવો

    b. There are four cells or categories in each of the new tables.

    df = number of cells – 1 = 4 – 1 = 3.

11. A factory manager needs to understand how many products are defective versus how many are produced. The number of expected defects is listed in .

    Number Produced	Number Defective
    0–100	5
    101–200	6
    201–300	7
    301–400	8
    401–500	10

    A random sample was taken to determine the actual number of defects. shows the results of the survey.

    Number Produced	Number Defective
    0–100	5
    101–200	7
    201–300	8
    301–400	9
    401–500	11

    State the null and alternative hypotheses needed to conduct a goodness-of-fit test, and state the degrees of freedom.

    જવાબ બતાવો

    H₀:The number of defaults fits expectations.

    H_a:The number of defaults does not fit expectations.
    df = 4

12. Employers want to know which days of the week employees are absent in a five-day work week. Most employers would like to believe that employees are absent equally during the week. Suppose a random sample of 60 managers were asked on which day of the week they had the highest number of employee absences. The results were distributed as in . For the population of employees, do the days for the highest number of absences occur with equal frequencies during a five-day work week? Test at a 5 percent significance level.

    	Monday	Tuesday	Wednesday	Thursday	Friday
    Number of Absences	15	12	9	9	15

    જવાબ બતાવો

    The null and alternative hypotheses are as follows:

    • H₀: The absent days occur with equal frequencies; that is, they fit a uniform distribution.
    • H_a: The absent days occur with unequal frequencies; that is, they do not fit a uniform distribution.

    If the absent days occur with equal frequencies, then, out of 60 absent days (the total in the sample: 15 + 12 + 9 + 9 + 15 = 60) there would be 12 absences on Monday, 12 on Tuesday, 12 on Wednesday, 12 on Thursday, and 12 on Friday. These numbers are the expected (E) values. The values in the table are the observed (O) values or data.

    This time, calculate the χ² test statistic by hand. Make a chart with the following headings and fill in the columns:

    • Expected (E) values (12, 12, 12, 12, 12)
    • Observed (O) values (15, 12, 9, 9, 15)
    • (O – E)
    • (O – E)²
    • \(\frac{{(O\text{ - }E)}^{2}}{E}\)

    Now add (sum) the last column. The sum is three. This is the χ² test statistic.

    To find the p-value, calculate P(χ² > 3). This test is right-tailed. Use a computer or calculator to find the p-value. You should get p-value = 0.5578.

    The dfs are the number of cells – 1 = 5 – 1 = 4.

    Next, complete a graph like the following one with the proper labeling and shading. You should shade the right tail.

    The decision is not to reject the null hypothesis.

    Conclusion: At a 5 percent level of significance, from the sample data, there is not sufficient evidence to conclude that the absent days do not occur with equal frequencies.

13. Teachers want to know which night each week their students are doing most of their homework. Most teachers think that students do homework equally throughout the week. Suppose a random sample of 56 students were asked on which night of the week they did the most homework. The results were distributed as in .

    	Sunday	Monday	Tuesday	Wednesday	Thursday	Friday	Saturday
    Number of Students	11	8	10	7	10	5	5

    From the population of students, do the nights for the highest number of students doing the majority of their homework occur with equal frequencies during a week? What type of hypothesis test should you use?

    જવાબ બતાવો

    df = 6

    p-value = 0.6093
    We decline to reject the null hypothesis. There is not enough evidence to support that students do not do the majority of their homework equally throughout the week.

14. At the 1 percent significance level, does it appear that the distribution number of televisions of far western U.S. families is different from the distribution for the American population as a whole?

    જવાબ બતાવો

    This problem asks you to test whether the far western U.S. families distribution fits the distribution of the American families. This test is always right-tailed.

    The first table contains expected percentages. To get expected (E) frequencies, multiply the percentage by 600. The expected frequencies are shown in .

    Number of Televisions	Percent	Expected Frequency
    0	10	(0.10)(600) = 60
    1	16	(0.16)(600) = 96
    2	55	(0.55)(600) = 330
    3	11	(0.11)(600) = 66
    more than 3	8	(0.08)(600) = 48

    Therefore, the expected frequencies are 60, 96, 330, 66, and 48. In the TI calculators, you can let the calculator do the math. For example, instead of 60, enter 0.10 * 600.

    H₀: The number of televisions distribution of far western U.S. families is the same as the number of televisions distribution of the American population.

    H_a: The number of televisions distribution of far western U.S. families is different from the number of televisions distribution of the American population.

    Distribution for the test: \({\chi }_{4}^{2}\) where df = (the number of cells) – 1 = 5 – 1 = 4.

    Calculate the test statistic: χ² = 29.65

    Graph

    Probability statement: p-value = P(χ² > 29.65) = .000006

    Compare α and the p-value:

    • α = .01
    • p-value = 0.000006

    So, α > p-value.

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

    This means you reject the hypothesis that the distribution for the far western states is the same as that of the American population as a whole.

    Conclusion: At the 1 percent significance level, from the data, there is sufficient evidence to conclude that the number of televisions distribution for the far western United States is different from the number of televisions distribution for the American population as a whole.

15. The expected percentage of the number of pets students have in their homes is distributed (this is the given distribution for the student population of the United States) as in .

    Number of Pets	Percent
    0	18
    1	25
    2	30
    3	18
    4+	9

    A random sample of 1,000 students from the eastern United States resulted in the data in .

    Number of Pets	Frequency
    0	210
    1	240
    2	320
    3	140
    4+	90

    At the 1 percent significance level, does it appear that the distribution number of pets of students in the eastern United States is different from the distribution for the United States student population as a whole? What is the p-value?

    જવાબ બતાવો

    p-value = 0.0036

    We reject the null hypothesis that the distributions are the same. There is sufficient evidence to conclude that the distribution for “number of pets” of students in the Eastern United States is different from the distribution for the U.S. student population as a whole.

16. Suppose you flip two coins 100 times. The results are 20 HH, 27 HT, 30 TH, and 23 TT. Are the coins fair? Test at a 5 percent significance level.

    જવાબ બતાવો

    This problem can be set up as a goodness-of-fit problem. The sample space for flipping two fair coins is {HH, HT, TH, TT}. Out of 100 flips, you would expect 25 HH, 25 HT, 25 TH, and 25 TT. This is the expected distribution. The question, “Are the coins fair?” is the same as saying, “Does the distribution of the coins (20 HH, 27 HT, 30 TH, 23 TT) fit the expected distribution?”

    Random variable: Let X = the number of heads in one flip of the two coins. X takes on the values 0, 1, 2. There are 0, 1, or 2 heads in the flip of two coins. Therefore, the number of cells is three. Since X = the number of heads, the observed frequencies are 20 for two heads, 57 for one head, and 23 for zero heads or both tails. The expected frequencies are 25 for two heads, 50 for one head, and 25 for zero heads or both tails. This test is right-tailed.

    H₀: The coins are fair.

    H_a: The coins are not fair.

    Distribution for the test: \({\chi }_{2}^{2}\) where df = 3 – 1 = 2.

    Calculate the test statistic: χ² = 2.14.

    Graph

    Probability statement: p-value = P(χ² > 2.14) = 0.3430.

    Compare α and the p-value:

    • α = .05
    • p-value = 0.3430

    α < p-value.

    Make a decision: Since α < p-value, do not reject H₀.

    Conclusion: There is insufficient evidence to conclude that the coins are not fair.

17. Students in a social studies class hypothesize that the literacy rates around the world for every region are 82 percent. shows the actual literacy rates around the world broken down by region. What are the test statistic and the degrees of freedom?

    MDG Region	Adult Literacy Rate (%)
    Developed regions	99
    Commonwealth of Independent States	99.5
    Northern Africa	67.3
    Sub-Saharan Africa	62.5
    Latin America and the Caribbean	91
    Eastern Asia	93.8
    Southern Asia	61.9
    Southeastern Asia	91.9
    Western Asia	84.5
    Oceania	66.4

    જવાબ બતાવો

    degrees of freedom = 9

    chi² test statistic = 26.38

    Press STAT and ENTER. Make sure you clear lists L1, L2, and L3 if they have data in them. Into L1, put the observed frequencies 99, 99.5, 67.3, 62.5, 91, 93.8, 61.9, 91.9, 84.5, 66.4. Into L2, put the expected frequencies 82, 82, 82, 82, 82, 82, 82, 82, 82, 82. Arrow over to list L3 and up to the name area "L3". Enter (L1-L2)^2/L2 and ENTER. Press 2nd QUIT. Press 2nd LIST and arrow over to MATH. Press 5. You should see "sum". Enter L3. Rounded to two decimal places, you should see 26.38. Press 2nd DISTR. Arrow down to 7:χ2cdf (or press 7). Press ENTER. Enter 26.38,1E99,9). Rounded to four places, you should see .0018, which is the p-value.

    The newer TI-84 calculators have in STAT TESTS the test Chi2 GOF. To run the test, put the observed values (the data) into a first list and the expected values (the values you expect if the null hypothesis is true) into a second list. Press STAT TESTS and Chi2 GOF. Enter the list names for the Observed list and the Expected list. Enter the degrees of freedom and press calculate or draw. Make sure you clear any lists before you start.

18. An archeologist is calculating the distribution of the frequency of the number of artifacts she finds in a dig site. Based on previous digs, the archeologist creates an expected distribution broken down by grid sections in the dig site. Once the site has been fully excavated, she compares the actual number of artifacts found in each grid section to see if her expectation was accurate.

19. An economist is deriving a model to predict outcomes on the stock market. He creates a list of expected points on the stock market index for the next two weeks. At the close of each day’s trading, he records the actual points on the index. He wants to see how well his model matched what actually happened.

    જવાબ બતાવો

    a goodness-of-fit test

20. A personal trainer is putting together a weight-lifting program for her clients. For a 90-day program, she expects each client to lift a specific maximum weight each week. As she goes along, she records the actual maximum weights her clients lifted. She wants to know how well her expectations met with what was observed.

21. \(df=\) ______

    જવાબ બતાવો

    3

22. State the null and alternative hypotheses.

23. χ² test statistic = ______

    જવાબ બતાવો

    2.04

24. p-value = ______

25. At the 5 percent significance level, what can you conclude?

    જવાબ બતાવો

    We decline to reject the null hypothesis. There is not enough evidence to suggest that the observed test scores are significantly different from the expected test scores.

26. If the ethnicities of patients followed the ethnicities of the total county population, fill in the expected number of cases per ethnic group.
    Perform a goodness-of-fit test to determine whether the occurrence of disease cases follows the ethnicities of the general population of Santa Clara County.

27. H₀: _______

    જવાબ બતાવો

    H₀: the distribution of disease cases follows the ethnicities of the general population of Santa Clara County.

28. H_a: _______

29. Is this a right-tailed, left-tailed, or two-tailed test?

    જવાબ બતાવો

    right-tailed

30. degrees of freedom = _______

31. χ² test statistic = _______

    જવાબ બતાવો

    2016.136

32. p-value = _______

33. Graph the situation. Label and scale the horizontal axis. Mark the mean and test statistic. Shade in the region corresponding to the p-value.

    Let α = 0.05.

    Decision: ________________

    Reason for the decision: ________________

    Conclusion (write out in complete sentences): ________________

    જવાબ બતાવો

    Graph: Check student’s solution.

    Decision: Reject the null hypothesis.

    Reason for decision: p-value < alpha

    Conclusion: The make-up of cases does not fit the ethnicities of the general population of Santa Clara County.

34. Does it appear that the pattern of disease cases in Santa Clara County corresponds to the distribution of ethnic groups in this county? Why or why not?

35. A six-sided die is rolled 120 times. Fill in the expected frequency column. Then, conduct a hypothesis test to determine if the die is fair. The data in are the result of the 120 rolls.

    Face Value	Frequency	Expected Frequency
    1	15
    2	29
    3	16
    4	15
    5	30
    6	15

36. The marital status distribution of the U.S. male population, ages 15 and older, is as shown in .

    Marital Status	%	Expected Frequency
    Never Married	31.3%
    Married	56.1%
    Widowed	2.5%
    Divorced/Separated	10.1%

    Suppose that a random sample of 400 U.S. males, 18 to 24 years old, yielded the following frequency distribution. We are interested in whether this age group of males fits the distribution of the U.S. adult population. Calculate the frequency one would expect when surveying 400 people. Fill in , rounding to two decimal places.

    Marital Status	Frequency
    Never Married	140
    Married	238
    Widowed	2
    Divorced/Separated	20

    જવાબ બતાવો

    Marital Status	%	Expected Frequency
    Never Married	31.3%	125.2
    Married	56.1%	224.4
    Widowed	2.5%	10
    Divorced/Separated	10.1%	40.4

    1. The data fit the distribution.
    2. The data do not fit the distribution.
    3. 3
    4. chi-square distribution with df = 3
    5. 19.27
    6. 0.0002
    7. Check student’s solution.
    8. 1. Alpha = 0.05
       2. Decision: Reject null hypothesis.
       3. Reason for decision: p-value < alpha
       4. Conclusion: Data do not fit the distribution.

37. Perform a goodness-of-fit test to determine whether the local results follow the distribution of the U.S. overall student population based on ethnicity.

38. Perform a goodness-of-fit test to determine whether the local results follow the distribution of U.S. AP examinee population, based on ethnicity.

    જવાબ બતાવો

    1. H₀: The local results follow the distribution of the U.S. AP examinee population.
    2. H_a: The local results do not follow the distribution of the U.S. AP examinee population.
    3. df = 5
    4. chi-square distribution with df = 5
    5. chi-square test statistic = 13.4
    6. p-value = 0.0199
    7. Check student’s solution.

    8. 1. Alpha = 0.05
       2. Decision: Reject null when a = 0.05.
       3. Reason for decision: p-value < alpha
       4. Conclusion: Local data do not fit the AP examinee distribution.
       5. Decision: Do not reject null when a = 0.01
       6. Conclusion: There is insufficient evidence to conclude that local data do not follow the distribution of the U.S. AP examinee distribution.

39. The city of South Lake Tahoe, California, has an Asian population of 1,419 out of a total population of 23,609. Suppose that a survey of 1,419 self-reported Asians in the borough of Manhattan in the New York City area yielded the data in . Conduct a goodness-of-fit test to determine if the self-reported subgroups of Asians in Manhattan fit that of the South Lake Tahoe area.

    Race	South Lake Tahoe Frequency	Manhattan Frequency
    Asian Indian	131	174
    Chinese	118	557
    Filipino	1,045	518
    Japanese	80	54
    Korean	12	29
    Vietnamese	9	21
    Other	24	66

40. Conduct a goodness-of-fit test to determine if the actual college majors of graduating females fit the distribution of their expected majors.

    Major	Females—Expected Major	Females—Actual Major
    Arts & Humanities	14%	670
    Biological Sciences	8.4%	410
    Business	13.1%	685
    Education	13%	650
    Engineering	2.6%	145
    Physical Sciences	2.6%	125
    Professional	18.9%	975
    Social Sciences	13%	605
    Technical	0.4%	15
    Other	5.8%	300
    Undecided	8%	420

    જવાબ બતાવો

    1. H₀: The actual college majors of graduating females fit the distribution of their expected majors.
    2. H_a: The actual college majors of graduating females do not fit the distribution of their expected majors.
    3. df = 10
    4. chi-square distribution with df = 10
    5. test statistic = 11.48
    6. p-value = 0.3211
    7. Check student’s solution.
    8. 1. Alpha = 0.05
       2. Decision: Do not reject null hypothesis when a = 0.05 and a = 0.01.
       3. Reason for decision: p-value > alpha
       4. Conclusion: There is insufficient evidence to conclude that the distribution of actual college majors of graduating females do not fit the distribution of their expected majors.

Symbols used here

\sqrt{x},\ \sqrt[n]{x}

square root, n-th root
The non-negative number whose square (n-th power) is x.

\sigma,\ s,\ \sigma^2

standard deviation, sample s.d., variance
Typical distance from the mean; its square.

i

imaginary unit
i² = −1.

\leq,\ \geq

less/greater than or equal
Inequalities that allow equality; < and > exclude it.

\pm

plus or minus
Both signs at once: x = 3 ± 2 means 5 and 1.

\approx

approximately equal
Equal to the precision shown, not exactly.

n!

factorial
n × (n−1) × … × 1; the number of orderings of n things. 0! = 1.

\binom{n}{k}

binomial coefficient, "n choose k"
Number of k-element subsets of n things: n!/(k!(n−k)!).

\sum_{k=1}^{n} a_k

summation
Add a_k for k = 1 up to n.

A \cup B,\ A \cap B,\ A \setminus B

union, intersection, difference
In either; in both; in A but not B.

\bar{x},\ \mu

sample mean, population mean
Average of the data; average of the whole population.

P(A),\ P(A \mid B)

probability, conditional probability
Chance of A; chance of A given that B happened.

E[X],\ \operatorname{Var}(X)

expected value, variance
Probability-weighted average of X; its spread.

N(\mu, \sigma^2),\ z

normal distribution, z-score
The bell curve with mean μ and variance σ²; (x − μ)/σ.

How to: Chi-square tests

1. Evaluate.

Questions people ask

Mean or median — which should I use?

Median when the data have outliers or a long tail (incomes, house prices); mean when the data are roughly symmetric and you want every value to count. Report both if they disagree — the gap is itself information.

What does a p-value actually say?

The probability of seeing data at least this extreme if the null hypothesis were true. It is not the probability that the null hypothesis is true.

Why divide by n − 1 for the sample variance?

The sample mean sits closer to the sample than the true mean does, so squared deviations from it are slightly too small on average; dividing by n − 1 instead of n corrects the bias.

તમારા પોતાના પ્રયત્ન કરો

⌨ Type ✎ Draw

તમારી ગણિતની સમસ્યા

ઉકેલો

x²√÷×π( )=∫d/dxlim[ ]sinlogeˣ

(18-20)^2/20 + (22-20)^2/20 + (20-20)^2/20(45-50)^2/50 + (55-50)^2/50(3-1)*(4-1)(10-15)^2/15 + (20-15)^2/15

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

આમાં વધુ Statistics & Probability

Sampling and dataDescribing data with graphsMean, median and modeProbabilityCounting: permutations and combinationsDiscrete random variablesContinuous random variablesThe normal distributionThe central limit theoremConfidence intervalsHypothesis testingComparing two samplesLinear regression and correlationANOVA and the F distribution