Question

5. A large supermarket carries four qualities of beef. Customers are believed to purchase these four qualities with probabilities of 0.10, 0.30, 0.35, and 0.25, respectively, from the least to most expensive. A sample of 500 purchases resulted in sales of 48, 162, 191, and 99 of respective qualities. Does this sample contradict the expected proportions? Use a) α = 0.05 b) α =0.025

Answer 1

Chi-Square Goodness of Fit test - 5%

(1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: H0: p1 =0.1, p2 =0.3, p3​=0.35, p4​=0.25 Ha​: Some of the population proportions differ from the values stated in the null hypothesis This corresponds to a Chi-Square test for Goodness of Fit. (2) Degrees of Freedom The number of degrees of freedom is df=n-1=4-1=3 (3) Test Statistics The Chi-Squared statistic is computed as follows: (4)Critical Value and Rejection Region Based on the information provided, the significance level is α=0.05, the number of degrees of freedom is df=n-1=4-1=3, so the critical value becomes 7.8147. Then the rejection region for this test is R={χ2:χ2>7.8147}. (5)P-value The P-value is the probability that a chi-square statistic having 3 degrees of freedom is more extreme than 7.8147. The p-value is p=Pr(χ2>7.9109)=0.0479 (6) The decision about the null hypothesis Since it is observed that χ2=7.9109>χ2_crit​=7.8147, it is then concluded that the null hypothesis is rejected. (7) Conclusion It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that some of the population proportions differ from those stated in the null hypothesis, at the α=0.05 significance level.

Chi-Square Goodness of Fit test - 2.5%

(1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: H0: p1 =0.1, p2 =0.3, p3​=0.35, p4​=0.25 Ha​: Some of the population proportions differ from the values stated in the null hypothesis This corresponds to a Chi-Square test for Goodness of Fit. (2) Degrees of Freedom The number of degrees of freedom is df=n-1=4-1=3 (3) Test Statistics The Chi-Squared statistic is computed as follows: (4)Critical Value and Rejection Region Based on the information provided, the significance level is α=0.025, the number of degrees of freedom is df=n-1=4-1=3, so the critical value becomes 9.3484. Then the rejection region for this test is R={χ2:χ2>9.3484}. (5)P-value The P-value is the probability that a chi-square statistic having 3 degrees of freedom is more extreme than 9.3484. The p-value is p=Pr(χ2>7.9109)=0.0479 (6) The decision about the null hypothesis Since it is observed that χ2=7.9109≤χc2​=9.3484, it is then concluded that the null hypothesis is not rejected. (7) Conclusion It is concluded that the null hypothesis Ho is not rejected. Therefore, there is NOT enough evidence to claim that some of the population proportions differ from those stated in the null hypothesis, at the α=0.025 significance level.