Question

A random sample is selected from a normal population with a mean of μ = 20 and a standard deviation of σ = 10. After a treatment is administered to the individuals in the sample, the sample mean is found to be M = 25. If the sample consists of n = 4 scores, is the sample mean sufficient to conclude that the treatment has a significant effect? Use a two-tailed test with alpha = .05.

Answer 1

Solution: The provided sample mean is X = 25 and the known population standard deviation is o = 10, and the sample size is n = 4. (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: = 20 Ho:= Ha: 11 20 This corresponds to a two-tailed test, for which a z-test for one mean, with known population standard deviation will be used. (2) Rejection Region Based on the information provided, the significance level is a = 0.05, and the critical value for a two-tailed test is ze = 1.96. The rejection region for this two-tailed test is R= {z : 21 > 1.96} (3) Test Statistics The Z-statistic is computed as follows: X 25 - 20 = 1 0/vn 10/14 (4) Decision about the null hypothesis Since it is observed that :| =1<% = 1.96, it is then concluded that the null hypothesis is not rejected. Using the P-value approach: The p-value is p = 0.3173, and since p = 0.3173 > 0.05, it is concluded that the null hypothesis is not rejected. (5) Conclusion It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population mean is different than 20, at the 0.05 significance level. HO