A student claims the average amount of time spent studying for an exam is less than or equal to 120 minutes. A random sample of 101 students is drawn from a normal distribution. The sample mean is 140 with a sample standard deviation of 20. Test the claim with a .01 level of significance and write your conclusion
Z = (mean1 - mean2)/standard error (SE) of difference between means
SEdiff = √(SEmean1^2 + SEmean2^2)
SEm = SD/√n
If 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/probability related to the Z score.
To test the claim that the average amount of time spent studying for an exam is less than or equal to 120 minutes, we can use a one-sample t-test.
Here are the steps to perform the hypothesis test:
Step 1: State the null and alternative hypotheses.
- Null hypothesis (H0): The average time spent studying is equal to or greater than 120 minutes (μ ≥ 120).
- Alternative hypothesis (Ha): The average time spent studying is less than 120 minutes (μ < 120).
Step 2: Determine the significance level.
The significance level (α) is given as 0.01 in this case.
Step 3: Compute the test statistic.
The test statistic for a one-sample t-test is given by:
t = (sample mean - population mean) / (sample standard deviation / √sample size)
We have:
sample mean (x̄) = 140
population mean (μ) = 120
sample standard deviation (s) = 20
sample size (n) = 101
t = (140 - 120) / (20 / √101)
t ≈ 2.527
Step 4: Determine the critical value.
Since the alternative hypothesis states that the average time is less than 120 minutes, it is a one-tailed test. We need to find the critical value for a one-tailed test with a significance level of 0.01 and degrees of freedom (df) = n - 1 = 101 - 1 = 100.
Using a t-table or a calculator, the critical value at a significance level of 0.01 and df = 100 is approximately -2.626.
Step 5: Make a decision and write a conclusion.
If the test statistic is less than the critical value, we reject the null hypothesis; otherwise, we fail to reject the null hypothesis.
In this case, since t (2.527) > critical value (-2.626), we fail to reject the null hypothesis.
Conclusion: There is not enough statistical evidence to support the claim that the average amount of time spent studying for the exam is less than or equal to 120 minutes at a significance level of 0.01.