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 ì ó Sample Score p Tails of Test
A
100.0
10.0
80
.05
1 (low predicted)

B
100.0
20.0
80
.01
2

C
74.3
11.8
80
.01
2

D
16.9
1.2
80
.05
1 (low predicted)

E
88.1
12.7
80
.05 2

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. However, you are lacking the n values for your samples.

To determine whether to reject the null hypothesis for each of the studies, you need to compare the sample score to the Z-score cutoff on the comparison distribution. Here's how you can do that:

(a) Z-score cutoff on the comparison distribution:
For a one-tailed test with an alpha level of 0.05, the Z-score cutoff is ±1.645.
For a one-tailed test with an alpha level of 0.01, the Z-score cutoff is ±2.33.
For a two-tailed test with an alpha level of 0.05, the Z-score cutoff is ±1.96.

(b) Z-score for the sample score:
To calculate the Z-score for the sample score, you need to use the formula: Z = (sample score - population mean) / standard deviation.

(c) Conclusion:
Based on the Z-score cutoff and the Z-score for the sample score, the conclusion is as follows:

Study A:
Z-score cutoff: ±1.645 (one-tailed test with an alpha level of 0.05)
Z-score for the sample score: (80 - 100) / 10 = -2.0
Conclusion: Since the sample score is below the Z-score cutoff, reject the null hypothesis.

Study B:
Z-score cutoff: ±2.33 (one-tailed test with an alpha level of 0.01)
Z-score for the sample score: (80 - 100) / 20 = -1.0
Conclusion: Since the sample score is below the Z-score cutoff, reject the null hypothesis.

Study C:
Z-score cutoff: ±2.33 (two-tailed test with an alpha level of 0.01)
Z-score for the sample score: (80 - 74.3) / 11.8 = 0.48
Conclusion: Since the absolute value of the Z-score for the sample score is less than the Z-score cutoff, do not reject the null hypothesis.

Study D:
Z-score cutoff: ±1.645 (one-tailed test with an alpha level of 0.05)
Z-score for the sample score: (80 - 16.9) / 1.2 = 50.92
Conclusion: Since the sample score is above the Z-score cutoff, reject the null hypothesis.

Study E:
Z-score cutoff: ±1.96 (two-tailed test with an alpha level of 0.05)
Z-score for the sample score: (80 - 88.1) / 12.7 = -0.64
Conclusion: Since the absolute value of the Z-score for the sample score is less than the Z-score cutoff, do not reject the null hypothesis.

Remember, these conclusions are based on the assumption that all populations are normally distributed.