14. Based on the information given for each of the following...

14. 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 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

Transcript

00:01 Hello students in this question based on the given information, we have to decide whether we have to reject the null hypothesis or not.

00:09 So here for the study a we have given mean mu is equal to 13 standard deviation sigma is equal to 2 sample score.

00:19 That means the value of x is equal to 15 and this is the level of significance that is alpha is equal to 0 .05 and the test is a one tailed test.

00:31 Now by using this first we will find the value of z.

00:34 So z score is given by x minus mu divided by sigma which is equal to 15 minus 13 divided by 2 and that is equal to 1.

00:45 So this is the value of z.

00:48 Then for the one tailed test we will get the critical value critical value is equal to z alpha which is equal to z 0 .05 and that is equal to 1 .645.

01:04 This is by using the z table or standard normal distribution table.

01:11 And now if we compare these two values that is the calculated z value with the critical value and we know that the decision rule is when the z value is greater than the critical value, then we reject the null hypothesis.

01:27 So here since z is equal to 1 is less than the critical value 1 .645.

01:36 So therefore we fail to reject the null hypothesis that is h naught.

01:46 Then in the next question, we have given a mean is equal to 13 standard deviation 2 sample score is 15 and significance level 0 .05 and the test is a two tailed test two tailed test.

02:03 So now first we will compute z score which is x minus mu divided by sigma that is equal to 15 minus 13 divided by 2 and that is equal to 1.

02:14 Then for the two tailed test now the critical value critical value is z alpha by 2 which is equal to z 0 .05 divided by 2 that is z 0 .025 and this is equal to 1 .96.

02:33 Now here also since the value of z is equal to 1 is less than the critical value.

02:40 So therefore we fail to reject the null hypothesis that is h naught.

02:49 Then for the next question, we have given this mean standard deviation and sample score significance level is 0 .01 and the test is a one tailed test one tailed test...

Question

Please give Ace some feedback