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.
a. The null hypothesis (Ho) is that users learn the package in a mean time of 40 hours or more. The alternate hypothesis (H1) is that users learn the package in less than a mean of 40 hours.
Ho: μ ≥ 40 (where μ is the population mean)
H1: μ < 40
b. The decision rule depends on the significance level (α) chosen for the hypothesis test. In this case, α = 0.05. Since this is a one-tailed test (looking for evidence of users learning in less than 40 hours), we will reject the null hypothesis if the test statistic falls in the critical region, which is determined by the critical value for the chosen significance level.
c. To compute the value of the test statistic, we need to calculate the t-score using the sample mean, population mean, sample size, and sample standard deviation.
t = (sample mean - population mean) / (sample standard deviation / sqrt(sample size))
= (39 - 40) / (5 / sqrt(100))
= -1 / (5 / 10)
= -1 / 0.5
= -2
Here, the sample mean is 39, the population mean is 40, the sample standard deviation is 5, and the sample size is 100.
d. To compute the p-value, we need to find the probability of obtaining a t-score more extreme than -2 (in the left tail of the t-distribution) when the null hypothesis is true. This can be done using a t-distribution table or a statistical software.
The p-value for a t-score of -2 (degrees of freedom = 99) in a one-tailed test is approximately 0.025.
e. Comparing the p-value (0.025) to the significance level (0.05), we can see that the p-value is less than the significance level. This means that there is sufficient evidence to reject the null hypothesis. Therefore, we can conclude that users learn the package in less than a mean of 40 hours.
Interpretation: Based on the sample data, there is evidence to suggest that the mean time to learn the new statistics program is less than 40 hours.
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.
Z<-1.25)=(1-.1056)=0.8944
P-value of 0.894<0.05
Reject?