Question

SE - Ppool(1-Ppool) Ppool (1-Ppool) C.l. => (P1-P2) z* SE Interpretation: 3. Flat Tire and Missed Class. A classic story involves four carpooling students who missed a test and gave an excuse a flat tire. On the makeup test, the instructor asked the students to identify the particular tire that went flat. If they really didn't have a flat tire, would they be able to identify the same tire? The author asked 41 other students to identify the tire they would select. The results are listed in the following table (except for one student who selected the spare). Use a 0.05 significance level to test the author's claim that the results fit a uniform distribution. Tire Left Front Right Front Left Rear Right Rear. Number Selected 11 15 Ho: p = 0 (The tires selected by students are not equally likely.) HA: P +0 (The tires selected by students are equally likely.) Hypothesis Test: (Expected count) E = Test Statistic: x = 60,-€7) + 0-8) + ... + 0x= E)? ...x = F) E2 EK Critical Value: (intersection of df and a) P-value: dfk - 1 = Conclusion: Decision

Answer 1

As we are testing here whether the tires selected by students are equally likely or not, therefore the expected frequency for each cell is computed here as:

= Total sample size / 4

= 41/4 = 10.25

The chi square test statistic here is computed as:

$\chi^2 = \sum \frac{(O_i - E_i)^2}{E_i} = \frac{0.75^2 + 4.75^2 + 2.25^2 + 4.25^2}{10.25}= 4.5122$

Therefore 4.5122 is the required test statistic value here.

The degrees of freedom here is computed as:
df = k - 1 = 3

Therefore 3 is the required degrees of freedom here.

The level of significance is given in the problem here as:

= 0,05

For 3 degrees of freedom, the p-value here is computed from chi square distribution tables here as:

$p = P(\chi^2_3 > 4.5122)= 0.21$

Therefore 0.21 is the required p-value here.

As the p-value here is 0.21 > 0.05 which is the level of significance, therefore the test is not significant here and we cannot reject the null hypothesis. Therefore Do not reject H0 is the required decision here.

As the test is not significant here, we dont have sufficient evidence here to reject the statement that the tires selected by students are equally likely