Question

A pizza shop claims its average home delivery time is 29 minutes. A sample of 30 deliveries had a sample average of 31.6 minutes. Assume the population standard deviation for the​ shop's deliveries is 5.95 minutes. Complete parts a and b below.

a. Is there support for the​ shop's claim using the criteria that the sample average of 31.6 minutes falls within the symmetrical interval that includes​ 95% of the sample means if the true population mean is 29 ​minutes? Select the correct choice below and fill in the answer box within your choice.

​(Type an integer or decimal rounded to four decimal places as​ needed.)

The probability P(X dash >= 31.6) = ______. This indicates that the sample average of 31.6 minutes falls outside the​ 95% symmetrical​ interval, contradicting the​ shop's claim.

Need help using excel.

Answer 1

Answer 2

To determine if there is support for the pizza shop's claim, we can perform a hypothesis test using the sample data.

Null Hypothesis (H0): The true population mean delivery time is 29 minutes. Alternative Hypothesis (Ha): The true population mean delivery time is not 29 minutes.

Given data: Sample size (n) = 30 Sample mean (x̄) = 31.6 minutes Population standard deviation (σ) = 5.95 minutes

a. To test the hypothesis, we can use the z-test since we know the population standard deviation and the sample size is large (n ≥ 30). The z-score can be calculated as follows:

$�=\frac{\overline{�}-�}{\frac{�}{\sqrt{�}}}$

where x̄ is the sample mean, μ is the population mean, σ is the population standard deviation, and n is the sample size.

Substitute the values:$�=\frac{31.6-29}{\frac{5.95}{\sqrt{30}}}$

Now calculate the z-score using Excel or a calculator:

$�\approx 2.23$

Next, we need to find the probability P(X̄ ≥ 31.6) given the z-score. We can use the standard normal distribution table or Excel's NORM.S.DIST function to find this probability. In Excel, you can use the formula:

$=1-����\mathrm{.}�\mathrm{.}����(2.23,����)$

This will give you the probability that the sample mean is greater than or equal to 31.6 minutes.

The result is approximately 0.0135 (rounded to four decimal places). Since this probability is less than 0.025 (which is half of the 0.05 significance level for a two-tailed test), we reject the null hypothesis. Therefore, there is support to conclude that the sample average of 31.6 minutes falls outside the 95% symmetrical interval, contradicting the pizza shop's claim.