Posted by **Michelle** on Monday, December 7, 2009 at 8:30pm.

A recent article in a computer magazine suggested that the mean time to fully learn a new software program is 40 hours. A sample of 100 first time users of a new statistics program revealed the mean time to learn it was 39 hours with the standard deviation of 5 hours. At the 0.05 significance level, can we conclude that users learn the package in less than a mean of 40 hours?

a. State the null and alternate hypotheses.

Ho:

H1:

b. State the decision rule.

c. Compute the value of the test statistic.

d. Compute the p-value.

e. What is your decision regarding the null hypothesis? Interpret the result.

- college Math -
**MathGuru**, Thursday, December 10, 2009 at 7:17pm
Ho: µ = 40 --->meaning the population mean is equal to 40.

H1: µ < 40 ---> meaning the population mean is less than 40.

This will be a one-tailed test because the alternative hypothesis is showing a specific direction.

Use a z-test. Find the significance level for a one-tailed test using a z-table. Compare the test statistic from the z-test to the value from the z-table. If the test statistic exceeds the value in the table, then reject the null and conclude µ < 40. The p-value is the actual level of the test statistic found using a z-table.

Here is a z-test formula to get you started:

z = (sample mean - population mean)/(standard deviation divided by the square root of the sample size)

Sample mean = 39

Population mean = 40

Standard deviation = 5

Sample size = 100

Fill the values into the formula and compute the test statistic.

I'll let you take it from here.

I hope this will help get you started.

- college Math -
**Karen**, Monday, March 7, 2016 at 5:29pm
Z<-1.25)=(1-.1056)=0.8944

P-value of 0.894<0.05

Reject?

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