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.
3 answers
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.
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.
2 answers