Based on the information given for each of the following studies, decide whether to reject the null hypothesis. For each, give (a) the Z-score cutoff (or cutoffs) on the comparison distribution at which the null hypothesis should be rejected, (b) the Z score on the comparison distribution for the sample score, and (c) your conclusion. Assume that all populations are normally distributed.
Population
Study μ σ Sample Score p Tails of Test
A 5 1 7 .05 1 (high predicted)
B 5 1 7 .05 2
C 5 1 7 .01 1 (high predicted)
D 5 1 7 .01 2
I would prefer not the answer given but please show me how to do it. I just do not understand.
What level of significance are you using? P ≤ .05? P ≤ .01?
Z = (score-mean)/SD
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportions related to the Z scores. For a one-tailed test, find Z with .05, but with a two-tailed test, look for Z with .025. If using P ≤ .01, use similar process.
i have the answers, but does not have the answers for the set of problems I was working on. That was one of my questions I was not completly sure of. I copied the problem word for word from the text. I did as much calculations as I knew how...it was then trying to figure out how to get the probability i think. This chapter threw me for a loop.
To determine whether to reject the null hypothesis for each study, we need to compare the sample score with the Z-score cutoff on the comparison distribution. The Z-score cutoff represents the critical value at which we would reject the null hypothesis. Let's go through each study step by step:
Study A:
Population:
Mean (μ) = 5
Standard deviation (σ) = 1
Given:
Sample score = 7
Alpha level (significance level) = 0.05
Tails of the test = 1 (high predicted)
(a) To find the Z-score cutoff, we need to identify the critical Z-score at the 0.05 significance level in a one-tailed test.
Using a Z-table or a statistical software, we can find that the critical Z-score for a one-tailed test at the 0.05 significance level is approximately 1.645.
(b) To find the Z-score for the sample score, we calculate it using the formula:
Z = (sample score - population mean) / population standard deviation
Z = (7 - 5) / 1
Z = 2
(c) To determine the conclusion, we compare the Z-score for the sample score with the Z-score cutoff. Since the Z-score for the sample score (2) is greater than the Z-score cutoff (1.645), we can reject the null hypothesis.
Conclusion: Based on the given data, we reject the null hypothesis for Study A.
You can follow the same steps for the remaining studies (B, C, and D) by using the provided information. Remember to calculate the Z-score cutoff, Z-score for the sample score, and compare them to make your conclusions.