Question

The mean waiting time at the drive-through of a fast-food restaurant from the time an order is placed to the time the order is received is 84.3 seconds. A manager devises a new drive-through system that he believes will decrease wait time, on average. He initiates the new system at his restaurant and measures the wait time for 12 randomly selected orders. Summary statistics for the 12 orders are given below: Mean 78.4 Standard deviation 14.1

Calculate the test-statistic (with degrees of freedom) to test the manager’s claim that the new drive-through system will decrease wait time, on average.

Show work. No work = no credit. Determine the p-value based on the test-statistic from part a. Based on your p-value from part b, state the conclusion in a complete sentence in the context of the problem.

Answer 1

Solution:-

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: u > 84.3
Alternative hypothesis: u < 84.3

Note that these hypotheses constitute a one-tailed test.  

Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method is a one-sample t-test.

Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).

SE = s / sqrt(n)

S.E = 4.07032
DF = n - 1

D.F = 11
t = (x - u) / SE

t = - 1.45

where s is the standard deviation of the sample, x is the sample mean, u is the hypothesized population mean, and n is the sample size.

The observed sample mean produced a t statistic test statistic of -1.45.

Thus the P-value in this analysis is 0.087.

Interpret results. Since the P-value (0.087) is greater than the significance level (0.01), we cannot reject the null hypothesis.

From the above test we have sufficient evidence in the favor of the claim that the new drive-through system will decrease wait time, on average.