college Math
posted by Michelle .
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 pvalue.
e. What is your decision regarding the null hypothesis? Interpret the result.

college Math 
MathGuru
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 onetailed test because the alternative hypothesis is showing a specific direction.
Use a ztest. Find the significance level for a onetailed test using a ztable. Compare the test statistic from the ztest to the value from the ztable. If the test statistic exceeds the value in the table, then reject the null and conclude µ < 40. The pvalue is the actual level of the test statistic found using a ztable.
Here is a ztest 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
Z<1.25)=(1.1056)=0.8944
Pvalue of 0.894<0.05
Reject?
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