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?
Ho: µ = 40 -->population mean = 40
Ha: µ < 40 -->population mean < 40
Using a one-sample z-test:
z = (sample mean - population mean)/(standard deviation divided by the square root of the sample size)
With your data:
z = (39 - 40)/(8/√100)
Finish the calculation. Determine whether or not to reject the null using 0.05 level of significance for a one-tailed test.
I hope this will help.
To determine if we can conclude that users learn the package in less than a mean of 40 hours, we can conduct a one-sample t-test. Here's how to calculate it:
Step 1: State the null and alternative hypotheses:
- Null Hypothesis (H₀): The mean time to learn the new statistics program is equal to 40 hours.
- Alternative Hypothesis (H₁): The mean time to learn the new statistics program is less than 40 hours.
Step 2: Set the significance level (alpha):
In this case, the significance level is given as 0.05.
Step 3: Compute the test statistic:
We will use the sample mean, the population mean, the sample standard deviation, and the sample size to calculate the test statistic.
Sample mean (x̄) = 39 hours
Population mean (μ) = 40 hours (given in the article)
Sample standard deviation (s) = 8 hours
Sample size (n) = 100
Using the formula for calculating the t-test statistic:
t = (x̄ - μ) / (s / √n)
Substituting the values into the formula:
t = (39 - 40) / (8 / √100)
t = -1 / 0.8
t = -1.25
Step 4: Determine the critical value:
Since the alternative hypothesis is one-sided (less than), we need to find the critical value corresponding to the lower tail at a significance level of 0.05 (given in the question). We can consult the t-distribution table or use statistical software. Let's assume the critical value for our calculation is -1.667.
Step 5: Compare the test statistic with the critical value:
If the test statistic (t-value) is less than the critical value, we can reject the null hypothesis.
In this case, -1.25 > -1.667, so we fail to reject the null hypothesis.
Step 6: Draw a conclusion:
Based on the analysis, at the 0.05 significance level, we do not have enough evidence to conclude that users learn the package in less than a mean of 40 hours.