Question

Recall the lunch at a fast-food restaurant problem from Assignment 4. Let µ represent the population mean amount spent for lunch ($) at a fast-food restaurant. Previously you calculated the mean and standard deviation of the fifteen sample measurements to be = $7.09 and s = $1.406, respectively. Suppose you want to determine if the true value of µ differs from $7.50.

a. Specify the null and alternative hypotheses for this test.

b. Since = $7.09 is less than $7.50, a manger wants to reject the null hypothesis. What are the problems with using such a decision rule?

c. Compute the value of the test statistic.

d. Find the approximate p-value of the test or use technology to find the exact p-value.

e. Select a value of α, the probability of a Type I error. What does α represent in the words of the problem.

f. Give the appropriate conclusion, based on the results of parts d and e.

g. What conditions must be satisfied for the test results to be valid? h. In Assignment 4, you found a 95% confidence interval for µ. Does this interval support your conclusion in part f?

Answer 1

a)

Hypotheses are:

b)

The sample mean can be by chance. It is quiet possible that in the next sample of same sample size sample mean is greater than population mean.

c)

Here we have following information:

So test statitics will be

d)

Degree of freedom:

df=n-1=14

The p-value using excel function "=TDIST(1.13,14,2)" is: 0.2775

e:

Let

It is known as level of signficance It shows the maximum allowable type I probability.

f)

Since p-value is greater than alpha so we fail to reject the null hypothesis.

That is we cannot conclude that the true value of µ differs from $7.50.

g)

The crticial values for 95% confidence interval using excel function "=TINV(0.05,14)" are 2.145

The confidence interval is:

yes because confidence interval contains $7.50.