A researcher hypothesizes that a new herbal supplement improves memory. A sample of n = 25 college students is obtained and each student takes the supplement daily for six weeks. At the end of the 6-week period, each student is given a standardized memory test, and the average score for the sample is M = 55. For the general population of college students, the distribution of test scores is normal with a mean of µ = 35 and a standard deviation of σ =10. Assume you're using a one-tailed test with alpha = .01.
1. What z-score marks the boundary for the critical region? Round your answer to the nearest tenth. Do not write anything else in the answer blank.
2. Do you reject or fail to reject the null hypothesis? Write either Reject or Fail to Reject. Write only one answer or the other, do not write anything else in the answer blank, and do not include periods or other punctuation.
1. FOR A ONE-TAILED TEST with alpha .01 = 2.33 (round accordingly)
2. Use a z-test to determine your test statistic.
z = (sample mean - population mean)/(standard deviation divided by the square root of the sample size)
Your data:
sample mean = 55
population mean = 35
standard deviation = 10
sample size = 25
Substitute the values above into the z-test and calculate. If the test statistic exceeds the critical value, then reject the null. If the test statistic does not exceed the critical value, fail to reject the null.
I hope this will help get you started.
To answer these questions, we need to find the critical z-score and compare it with the test statistic.
1. The critical region in this case is a one-tailed test, meaning we are only interested in extreme values in one direction. Since we are testing if the herbal supplement improves memory, we are looking for higher average scores. Because the mean for the general population is 35 and the sample mean is 55, we want to find the z-score that corresponds to a probability of 0.01 (alpha).
To find this z-score, we can use a z-table or a statistical software. For an alpha of 0.01, the z-score marking the boundary for the critical region is approximately 2.33 (rounding to the nearest tenth).
2. To know whether to reject or fail to reject the null hypothesis, we need to compare the test statistic (z-score) with the critical value. The test statistic is calculated by subtracting the population mean from the sample mean and dividing it by the standard error of the mean.
In this case, the null hypothesis assumes that the herbal supplement does not improve memory, so the population mean and the sample mean should be the same. The z-score can be calculated as follows:
z = (sample mean - population mean) / (standard deviation / square root of sample size)
Given:
Sample mean (x̄) = 55
Population mean (µ) = 35
Standard deviation (σ) = 10
Sample size (n) = 25
z = (55 - 35) / (10 / √25)
z = 20 / (10 / 5)
z = 20 / 2
z = 10
Comparing the test statistic (z = 10) with the critical value (2.33), we can see that the test statistic falls in the critical region (10 > 2.33).
Therefore, we reject the null hypothesis.