Question

The manager of a supermarket would like the variance of the waiting times of the customers not to exceed 3.7 minutes-squared. She would add a new cash register if the variance exceeds this threshold. She regularly checks the waiting times of the customers to ensure that the variance does not rise above the allowed level. In a recent random sample of 35 customer waiting times, she computes the sample variance as 5.8 minutes-squared. She believes that the waiting times are normally distributed. (You may find it useful to reference the appropriate table: chi-square table or F table)

a. Select the null and the alternative hypotheses to test if the threshold has been crossed. H0: σ2 ≤ 3.7; HA: σ2 > 3.7 H0: σ2 = 3.7; HA: σ2 ≠ 3.7 H0: σ2 ≥ 3.7; HA: σ2 < 3.7 b-1.

Calculate the value of the test statistic. (Round intermediate calculations to at least 4 decimal places and final answer to 2 decimal places.)

b-2. Find the p-value. 0.05 p-value < 0.10 p-value < 0.01 p-value 0.10 0.01 p-value < 0.025 0.025 p-value < 0.05 b-3. Do you reject the null hypothesis at the 5% significance level? Yes, since the p-value is less than the significance level. No, since the p-value is greater than the significance level. Yes, since the p-value is greater than the significance level. No, since the p-value is less than the significance level.

c. Is action required from the manager? No, since we do not reject the null hypothesis. Yes, since we do not reject the null hypothesis. No, since we reject the null hypothesis. Yes, since we reject the null hypothesis.

Answer 1

testing hypothesis to check whether the variance exceeds 3.7 or not.

(A) Hypotheses are

test statistic calculation

(B) p-value = CHISQ.DIST.RT(test statistic, n-1)

= CHISQ.DIST.RT(53.30,35-1)

= 0.0187 < 0.05, we will reject the null hypothesis

0.01 < p-value < 0.025

Yes, since the p-value is less than the significance level

(C) Yes, since we reject the null hypothesis.