Problem: A college bookstore tells prospective students that the average cost of

Question

Problem:

A college bookstore tells prospective students that the average cost of its textbooks is $72. A group of skeptical statistics students from the school thinks that the average cost is higher. In order to test the bookstore’s claim, the students will select a random sample of size n= 80 books. The mean from their random sample (x-bar) is $76.50 and standard deviation is $6.00. a.) (4 points) In your Excel file, state the appropriate null (H0) and alternative (Ha) hypotheses. You can write "Mu" for the population mean. b.) (4 points) In your Excel file, calculate the test statistic, or z-score, for the test. c.) (4 points) In your Excel file, use either a rejection region approach or a p-value approach to conclude on your test. State your conclusion statistically (reject or fail-to-reject Ho) and then restate the conclusion in terms of the actual problem.

Answer Ho : miu = 72 Ha : miu > 72 Z = ( 76.50 - 72 ) / ( 6 / srqt80 ) Z = 6.71 p value : 0.0000 since p value is lower than alpha ( 0.05) we reject Ho conclusion there is sufficeint evidence to support the claim that average cost is hihger than 72

My question: How did you find alpha?

Explanation / Answer

Solution:

if nothing is mentioned we take the alpha=0.05

In the above problem they didnt mention the confidence interval we take it 95%

alpha=level of significance=1-CI=-1-95%=1-0.95=0.05

This is how we got alpha=0.05

since Z calculated =6.71

sample size =n=80(large sample apply z test)

z critiical value for 95% is 1.96(right tail test)

since z calculated> z critical for 95% CI

Decision:Reject Null Hypothesis

using p value approach:

p value for z =6.71 at 0.05 significance level and for one tail test is

The P-Value is < 0.00001.

since  0.00001< alpha

that is  0.00001<0.05

The result is significant at p < 0.05.

Statistical conclusion:Reject Null Hypothesis.

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