Question

Deli Delivery delivers sandwiches to neighboring office buildings during lunch time in New York City. The deli claims that the sandwiches will be delivered within 20 minutes from receiving the order. Given the hectic schedules of their customers, consistent delivery time is a must. The owner has decided that the standard deviation of delivery times should be at most 6 minutes. To determine how consistently the sandwiches are being delivered, the manager randomly selects 29 orders and measures the time from receiving the order to delivery of the sandwich. The average time to delivery of the sample was 17 minutes with a standard deviation of 14.95 minutes. Will the manager conclude at α = 0.05 that the delivery times vary more than the owner desires?

Step 1 of 5:

Calculate the test statistic for this problem?

Step 2 of 5:

Calculate the p-value for an upper one-sided alternative hypothesis (right-tailed test). The alternative hypothesis for this problem is H₁:σ² > 36mins.

Step 3 of 5:

What is the rejection region at the 0.05 level of significance for the test on the variance for a lower-tailed alternative hypothesis (left-tail test)? The alternative hypothesis for this problem is H₁:σ² < 36mins.

Step 4 of 5:

State your conclusion using a significance level of 0.05 . The alternative hypothesis for this problem is H₁: σ² > 36 .

Step 5 of 5:

Calculate a 98% confidence interval for the variance of delivery times. Round your answers to four decimal places.

Answer 1

Solution:

We are given n = 29, Xbar = 17, S = 14.95, σ = 6, α = 0.05

Step 1 of 5:

Calculate the test statistic for this problem?

Here, we have to use Chi square test for population variance.

The test statistic formula is given as below:

Chi square = (n – 1)*S^2/σ^2

Chi square = (29 - 1)*14.95^2/6^2

Chi square = 173.835

Test statistic = 173.835

Step 2 of 5:

Calculate the p-value for an upper one-sided alternative hypothesis (right-tailed test). The alternative hypothesis for this problem is H₁:σ² > 36mins.

WE have

n = 29

df = n – 1 = 29 – 1 = 28

P-value = 0.0000

(by using Chi square table or excel)

Step 3 of 5:

What is the rejection region at the 0.05 level of significance for the test on the variance for a lower-tailed alternative hypothesis (left-tail test)? The alternative hypothesis for this problem is H₁:σ² < 36mins.

WE have

n = 29

df = n – 1 = 29 – 1 = 28

α = 0.05

Test is lower tailed.

Critical value = 16.9279

(by using Chi square table or excel)

Rejection region: Reject H₀ if test statistic Chi square < 16.9279

Step 4 of 5:

State your conclusion using a significance level of 0.05 . The alternative hypothesis for this problem is H₁: σ² > 36 .

We have

P-value = 0.0000

α = 0.05

P-value < α = 0.05

So, we reject the null hypothesis

There is sufficient evidence to conclude that population standard deviation is greater than 6.

Step 5 of 5:

Calculate a 98% confidence interval for the variance of delivery times. Round your answers to four decimal places.

Confidence interval for population variance

(n – 1)*S² / χ²_{α/2, n – 1} < σ² < (n – 1)*S² / χ²_{1 -α/2, n– 1}

We are given

Confidence level = 98%

Sample size = n = 29

Sample variance = S² = 14.95^2 = 223.5025

χ²_{α/2, n – 1} = 48.2782

χ²_{1 -α/2, n– 1} = 13.5647

(By using chi square table)

(n – 1)*S² / χ²_{α/2, n – 1} < σ² < (n – 1)*S² / χ²_{1 -α/2, n– 1}

(29 – 1)* 223.5025/48.2782 < σ² < (29 – 1)* 223.5025/13.5647

129.6251 < σ² < 461.3493

Lower limit = 129.6251

Upper limit = 461.3493