Question

they survey a sample of n= 25 music major students. participants respond to a measure of creativity and the average score for the sample is M=64. it is known that the distribution of creativity scores for the population of college students is normal with m=53. if the standard deviation is 40, it is the result sufficient to conclude that there is a significant difference?

Fail to reject or reject null hypothesis?

significant difference in creativity score or no significant difference ?

Answer 1

Step 1:

Ho: = 53

Ha: 53

Null hypothesis states that the average creativity score is 53

Alterantive hypothesis states that the average creativity score is not equal to 53.

Step 2:

n = 25

sample mean = 64

population sd = 40

Assuming that the data is normally distributed , also as the population sd is given, we will calculate z stat

Step 3:

As the level of significance is not given, we will take 0.05

The z-critical values for a two-tailed test, for a significance level of α=0.05

zc = −1.96 and zc = 1.96

As the z stat does not fall in the rejection area, we fail to reject the Null hypothesis.

Hence there is no sufficient evidence to believe tbat the creativity scores are different that population score of 53.