2. The director of admissions at a large university advises students that the cost of textbooks will average more than $300 for a typical semester. A random sample of 80 students enrolled in the university shows a sample mean cost of $315.40 and a sample standard deviation of $43.20. Use a 5% level of significance to test the director's claim.

I am wondering what the standard deviation is: is that the standard deviation on the textbook cost, or the standard deviation on the 80 students individual average?

standard deviation on the 80 students individual average

Determine your null and alternate hypothesis.

Use the z-test formula to find the test statistic:
z = (sample mean - population mean)/(standard deviation divided by the square root of the sample size)
z = (315.40 - 300)/(43.20/√80)

I'll let you finish the calculation.

Once you have the z-value from the above calculation, compare it to the critical value at .05 level of significance for a one-tailed test using a z-table. If the test value exceeds the critical value from the table, reject the null. If it does not exceed the critical value, do not reject the null. You can then draw your conclusions.

Hope this helps get you started.

To test the director's claim, we will conduct a hypothesis test using the provided information.

Step 1: State the hypotheses:
- Null hypothesis (H0): The average cost of textbooks is $300 or less.
- Alternative hypothesis (Ha): The average cost of textbooks is more than $300.

Step 2: Determine the level of significance (α):
The given question states a 5% level of significance, which means α = 0.05.

Step 3: Identify the test statistic:
Since we are dealing with sample means and the population standard deviation is unknown, we will use the t-test statistic.

Step 4: Formulate the decision rule:
- For a one-tailed test at a 5% significance level:
- The critical value for the right-tailed test would be found using α = 0.05 and degrees of freedom (df) = n - 1, where n is the sample size.

Step 5: Calculate the test statistic:
Given that the sample mean cost of textbooks is $315.40 and the sample standard deviation is $43.20, we can calculate the t-value using the formula:

t = (sample mean - hypothesized mean) / (sample standard deviation / sqrt(n))

where n is the sample size.

Substituting the values:
t = (315.40 - 300) / (43.20 / sqrt(80))

Step 6: Make a decision:
- If the calculated t-value falls in the critical region, we reject the null hypothesis.
- If the calculated t-value does not fall in the critical region, we fail to reject the null hypothesis.

Step 7: Calculate the critical value and compare it with the calculated t-value to make a decision.

Based on the sample of 80 students and a 5% level of significance, the degrees of freedom (df) would be 80 - 1 = 79.

Using a t-table or a statistical software, we determine the critical t-value at a 5% significance level (one-tailed test) with 79 degrees of freedom.

Step 8: Compare the critical value with the calculated t-value
If the calculated t-value is greater than the critical value, we reject the null hypothesis; if it is less, we fail to reject the null hypothesis.

Step 9: State the conclusion
Based on the decision in Step 8, we state our conclusion and interpret it in the context of the problem.

Note: Since the exact critical t-value at df = 79 cannot be determined without a t-table or statistical software, you would need to consult these resources to complete Steps 7 and 8.