Question

At Rochester Institute of Technology, 34% of the students are female. The Department of Mathematics and Statistics would like to know if the Data Analysis course has a different percentage of female students. From a random sample of 50 students taking this course, 23 were female. Does this data support that the proportion of females taking this class is different from the percentage of females in the school?

The appropriate hypotheses would be:

A) H o μ = 34 a n d H A μ ≠ 34

B) H o p = 0.34 a n d H A p ≠ 0.34

C) H o p ≤ 0.34 a n d H A p > 0.34

D) H o p ≥ 0.34 a n d H A p < 0.34

Answer: A, B, C, D?

The Test Statistic would be:?

The P-value would be:?   

The Conclusion would be:?

Answer 1

Solution:

Given:

At Rochester Institute of Technology, 34% of the students are female.

thus p = 0.34

Sample size = n = 50

x = 23

We have to test if this data support that the proportion of females taking this class is different from the percentage of females in the school.

that is:

The appropriate hypotheses would be

B) H_o : p = 0.34 and  H_A : p ≠ 0.34

The Test Statistic would be:........?

where

thus

The P-value would be:?   

For two tailed test , P-value is:

P-value = 2* P(Z > z test statistic) if z is positive

P-value = 2* P(Z < z test statistic) if z is negative

thus

P-value = 2* P(Z > z test statistic)

P-value = 2* P(Z >1.79 )

P-value = 2* [ 1 - P(Z <1.79 ) ]

Look in z table for z = 1.7 and 0.09 and find corresponding area.

P( Z< 1.79 ) = 0.9633

thus

P-value = 2* [ 1 - P(Z <1.79 ) ]

P-value = 2* [ 1 -0.9633 ]

P-value = 2* 0.0367

P-value = 0.0734

The Conclusion would be:?

Decision Rule:
Reject null hypothesis H0, if P-value < 0.05 level of significance, otherwise we fail to reject H0

Since P-value = 0.0734 > 0.05 level of significance, we fail to reject H0.

Thus at 0.05 level of significance, this data does not support that the proportion of females taking this class is different from the percentage of females in the school.