Question

In: Statistics and Probability

Each person in a representative sample of 446 college students age 18 to 24 was classified...

Each person in a representative sample of 446 college students age 18 to 24 was classified according to age and to the response to the following question: "How often have you used a credit card to buy items knowing you wouldn't have money to pay the bill when it arrived?" Possible responses were never, rarely, sometimes, or frequently. The responses are summarized in the table.

Age 18 to 20 Age 21 to 22 Age 23 to 24
Never 73 62 29
Rarely 35 34 32
Sometimes 31 42 40
Frequently 12 24 32

Do these data provide evidence that there is an association between age group and the response to the question? Test the relevant hypotheses using α = 0.01.

State the appropriate null and alternative hypotheses.

H0: There is no association between age group and the response to the question.
Ha: There is an association between age group and the response to the question.

H0: The proportions falling into each of the three age groups are the same for all four responses to the question.
Ha: The proportions falling into each of the three age groups are not the same for all four responses to the question.

H0: There is an association between age group and the response to the question.
Ha: There is no association between age group and the response to the question.

H0: The proportions falling into each of the three age groups are not the same for all four responses to the question.
Ha: The proportions falling into each of the three age groups are the same for all four responses to the question.

Find the test statistic and P-value. (Use technology. Round your test statistic to three decimal places and your P-value to four decimal places.)

X2=

P-value=

State the conclusion in the problem context.

Reject H0. There is convincing evidence that there is an association between age group and the response to the question.

Reject H0. There is not convincing evidence that there is an association between age group and the response to the question.

Fail to reject H0. There is not convincing evidence that there is an association between age group and the response to the question.

Fail to reject H0. There is convincing evidence that there is an association between age group and the response to the question.

Solutions

Expert Solution

Solution:

Here, we have to use chi square test for independence of two categorical variables.

Null hypothesis: H0: There is no association between age group and the response to the question.

Alternative hypothesis: Ha: There is an association between age group and the response to the question.

We are given level of significance = α = 0.01

Test statistic formula is given as below:

Chi square = ∑[(O – E)^2/E]

Where, O is observed frequencies and E is expected frequencies.

E = row total * column total / Grand total

We are given

Number of rows = r = 4

Number of columns = c = 3

Degrees of freedom = df = (r – 1)*(c – 1) = 3*2 = 6

α = 0.01

Critical value = 16.81189

(by using Chi square table or excel)

Calculation tables for test statistic are given as below:

Observed Frequencies

Column variable

Row variable

C1

C2

C3

Total

R1

73

62

29

164

R2

35

34

32

101

R3

31

42

40

113

R4

12

24

32

68

Total

151

162

133

446

Expected Frequencies

Column variable

Row variable

C1

C2

C3

Total

R1

55.52466

59.56951

48.90583

164

R2

34.19507

36.6861

30.11883

101

R3

38.25785

41.04484

33.69731

113

R4

23.02242

24.69955

20.27803

68

Total

151

162

133

446

Calculations

(O - E)

17.47534

2.430493

-19.9058

0.804933

-2.6861

1.881166

-7.25785

0.955157

6.302691

-11.0224

-0.69955

11.72197

(O - E)^2/E

5.500031

0.099166

8.102144

0.018948

0.196672

0.117494

1.376877

0.022228

1.178845

5.277194

0.019813

6.776037

Chi square = ∑[(O – E)^2/E] = 28.68545

Chi square = 28.685

P-value = 0.0001

(By using Chi square table or excel)

P-value < α = 0.01

So, we reject the null hypothesis

Reject H0. There is convincing evidence that there is an association between age group and the response to the question.

There is sufficient evidence to conclude that there is a association between age group and the response to the question.


Related Solutions

A college statistics instructor claims that the mean age of college statistics students is 24. A...
A college statistics instructor claims that the mean age of college statistics students is 24. A random sample of 116 college statistics students revealed a mean age of 22.7. The population standard deviation is known to be 5.68 years. Test his claim at the 0.05 level of significance. State the hypotheses and identify the claim. Find the critical value(s) Compute the test value. Make the decision to reject or not reject the null hypothesis. Summarize the results.
A college research group reported that 43​% of college students aged​ 18-24 would spend their spring...
A college research group reported that 43​% of college students aged​ 18-24 would spend their spring breaks relaxing at home in 2009. A sample of 165 college students was selected. Complete parts a through d below. a. Calculate the standard error of the proportion. σp = ​(Round to four decimal places as​ needed.) b. What is the probability that less than​ 40% of the college students from the sample spent their spring breaks relaxing at​ home? ​ P(Less than​ 40%...
The average GPA of a random sample of 18 college students who take evening classes was...
The average GPA of a random sample of 18 college students who take evening classes was calculated to be 2.94 with a standard deviation of 0.04. The average GPA of a random sample of 12 college students who take daytime classes was calculated to be 2.89 with a standard deviation of 0.05. Test the claim that the mean GPA of night students is larger than the mean GPA of day students at the .01 significance level. Claim: Select an answer...
Most students attending Faulkner University are older than the typical college entry age of 18. HRM...
Most students attending Faulkner University are older than the typical college entry age of 18. HRM students tend to be even older than the Faulkner average. Your own class of 17 students has an average age of 31.7, with a standard deviation of 6.7. If the average age of all students attending Faulkner at all campuses is 25.25, is your class significantly older? When I mean significantly older, I mean at the 95% level. Sample mean of your class’ age...
How old are college students? The national age distributions for college students are shown below. The...
How old are college students? The national age distributions for college students are shown below. The Western Association of Mountain Colleges took a random sample of 212 students and obtained the sample distribution listed second. Is the sample age distribution for the Western Association of Mountain Colleges a good fit to the national distribution? Use α = 0.05 National Age Distribution for College Students Age Under 26 26-35 36-45 46-55 Over 55 Percentage 39% 25% 16% 12% 8% Sample Distribution,...
At Rio Salado College, it was observed that 34% of the students were still classified as...
At Rio Salado College, it was observed that 34% of the students were still classified as dependents on their parents. However, in the honors program for students, 172 out of 418 students are dependents. The administrators want to know if the proportion of dependent students in the honors program is significantly different from the proportion for the school district. Test at the α=.05 level of significance. What is the hypothesized population proportion for this test? p= (Report answer as a...
At Rio Salado College, it was observed that 26% of the students were still classified as...
At Rio Salado College, it was observed that 26% of the students were still classified as dependents on their parents. However, in the honors program for students, 70 out of 222 students are dependents. The administrators want to know if the proportion of dependent students in the honors program is significantly different from the proportion for the school district. Test at the α=.05 level of significance. What is the hypothesized population proportion for this test? p= (Report answer as a...
The age distribution of students at a community college is given below. Age (years) Number of...
The age distribution of students at a community college is given below. Age (years) Number of Students (f) Under 21 416 21-25 420 26-30 219 31-35 51 Over 35 24 Total = 1130 Number of students (f) 416 420 219 51 24 1130 A student from the community college is selected at random. Find the conditional probability that the student is at most 35 given that he or she is at least 26.
For a woman age 18 to 24, systolic blood pressure’s (in mm of Hg) are normally...
For a woman age 18 to 24, systolic blood pressure’s (in mm of Hg) are normally distributed with a mean of 114.8 and a standard deviation of 13.1 (based on the data from the national health survey). A. If a woman between the ages of 18 and 24 is randomly selected, find the probability that her systolic blood pressure is above 120. B. If 30 women in that age bracket are randomly selected, find the probability that the main systolic...
Students in a representative sample of 65 first-year students selected from a large university in England...
Students in a representative sample of 65 first-year students selected from a large university in England participated in a study of academic procrastination. Each student in the sample completed the Tuckman Procrastination Scale, which measures procrastination tendencies. Scores on this scale can range from 16 to 64, with scores over 40 indicating higher levels of procrastination. For the 65 first-year students in this study, the mean score on the procrastination scale was 36.9 and the standard deviation was 6.46. (a)...
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT