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 8 hours. At the 0.05 significance level, can we conclude that users learn the package in less than a mean of 40 hours?
µ (population mean): _______
ó (population standard deviation): _______
X (sample mean): _______
s (sample standard deviation): _______
á (significance level): _______
ð (population proportion): _______
p (sample proportion): _______
a. State the null and alternate hypotheses.
H0: __________________
H1: __________________
b. Find the critical value
c. State the decision rule.
d. Compute the value of the test statistic.
e. Find the p-value
f. What is your decision regarding the null hypothesis? Interpret the result.
We do not do your work for you. Although it might require more time and effort, you will learn more if you do your own work. Isn't that why you go to school? Once you have answered your questions, we will be happy to give you feedback on your work.
However, here is some help.
The population standard deviation is not available from your data, but some of the other data is readily available from the problem statement.
Z = (mean1 - mean2)/standard error (SE) of difference between means
SEdiff = √(SEmean1^2 + SEmean2^2)
SEm = SD/√(n-1)
Since only one SD is provided, you can use just that to determine SEdiff.
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion related to your Z-score.
a. State the null and alternate hypotheses.
H0: The mean time to fully learn the new statistics software program is 40 hours.
H1: The mean time to fully learn the new statistics software program is less than 40 hours.
b. Find the critical value.
To find the critical value, we need to determine the level of significance, which is given as 0.05. Since we are testing whether the mean time is less than 40 hours, this is a left-tailed test. Looking up the critical value for a left-tailed test at a 0.05 significance level in a t-table with 99 degrees of freedom, the critical value is approximately -1.660.
c. State the decision rule.
If the value of the test statistic is less than the critical value, we will reject the null hypothesis. Otherwise, we will fail to reject the null hypothesis.
d. Compute the value of the test statistic.
The formula to compute the test statistic for this situation is:
t = (X - µ) / (s / sqrt(n))
where X is the sample mean (39 hours), µ is the population mean (40 hours), s is the sample standard deviation (8 hours), and n is the sample size (100).
Plugging in the values, we get:
t = (39 - 40) / (8 / sqrt(100))
t = -1 / (8 / 10)
t = -1.25
e. Find the p-value.
To find the p-value, we need to compare the test statistic (-1.25) to the t-distribution with 99 degrees of freedom. The p-value is the probability of obtaining a test statistic as extreme as the observed one, assuming the null hypothesis is true.
Looking up the p-value for -1.25 in a t-table with 99 degrees of freedom, the p-value is approximately 0.106.
f. What is your decision regarding the null hypothesis? Interpret the result.
Comparing the p-value (0.106) to the significance level (0.05), we can see that the p-value is greater than the significance level. Therefore, we fail to reject the null hypothesis.
This means that, based on the sample data, there is not enough evidence to conclude that users learn the statistics package in less than a mean of 40 hours.