In a representative sample of Penn State staff members there were 18 females and

Question

In a representative sample of Penn State staff members there were 18 females and 16 males. In a representative sample of Penn State faculty members there were 14 females and 20 males.

A. Construct a 95% confidence interval to estimate the difference in the proportion of Penn State staff who are female and the proportion of Penn State faculty who are female. If assumptions are met, use the normal approximation method. Do not do any calculations by hand. Use Minitab Express and remember to copy+paste all relevant output.

B. Use the five-step hypothesis testing procedure to determine if there is evidence that the Penn State faculty and Penn State staff groups differ in terms of the proportion of females. If assumptions are met, use the normal approximation method. Do not do any calculations by hand. Use Minitab Express and remember to copy+paste all relevant output.

Step 1: Check assumptions and write hypotheses

Step 2: Calculate the test statistic

Step 3: Determine the p-value

Step 4: Decide between the null and alternative hypotheses

Step 5: State a real world conclusion

Explanation / Answer

PART A.
TRADITIONAL METHOD
given that,
Penn State staff who are female = sample one, x1 =18, n1 =34, p1= x1/n1=0.5294
Penn State faculty who are female = = sample two, x2 =14, n2 =34, p2= x2/n2=0.4118
I.
standard error = sqrt( p1 * (1-p1)/n1 + p2 * (1-p2)/n2 )
where
p1, p2 = proportion of both sample observation
n1, n2 = sample size
standard error = sqrt( (0.5294*0.4706/34) +(0.4118 * 0.5882/34))
=0.1202
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, ? = 0.05
from standard normal table, two tailed z ?/2 =1.96
margin of error = 1.96 * 0.1202
=0.2356
III.
CI = (p1-p2) ± margin of error
confidence interval = [ (0.5294-0.4118) ±0.2356]
= [ -0.118 , 0.3533]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample one, x1 =18, n1 =34, p1= x1/n1=0.5294
sample two, x2 =14, n2 =34, p2= x2/n2=0.4118
CI = (p1-p2) ± sqrt( p1 * (1-p1)/n1 + p2 * (1-p2)/n2 )
where,
p1, p2 = proportion of both sample observation
n1,n2 = size of both group
a = 1 - (confidence Level/100)
Za/2 = Z-table value
CI = confidence interval
CI = [ (0.5294-0.4118) ± 1.96 * 0.1202]
= [ -0.118 , 0.3533 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 95% sure that the interval [ -0.118 , 0.3533] contains the difference between
true population proportion P1-P2
2) if a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the difference between
true population mean P1-P2

Minitab::
Test and CI for Two Proportions

Sample X N Sample p
1 18 34 0.529412
2 14 34 0.411765

Difference = p (1) - p (2)
Estimate for difference: 0.117647
95% CI for difference: (-0.117968, 0.353263)

PART B.
Given that,
sample one, x1 =18, n1 =34, p1= x1/n1=0.529
sample two, x2 =14, n2 =34, p2= x2/n2=0.412
null, Ho: p1 = p2
alternate, H1: p1 != p2
level of significance, ? = 0.05
from standard normal table, two tailed z ?/2 =1.96
since our test is two-tailed
reject Ho, if zo < -1.96 OR if zo > 1.96
we use test statistic (z) = (p1-p2)/?(p^q^(1/n1+1/n2))
zo =(0.529-0.412)/sqrt((0.471*0.529(1/34+1/34))
zo =0.972
| zo | =0.972
critical value
the value of |z ?| at los 0.05% is 1.96
we got |zo| =0.972 & | z ? | =1.96
make decision
hence value of |zo | < | z ? | and here we do not reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 0.9718 ) = 0.3311
hence value of p0.05 < 0.3311,here we do not reject Ho
ANSWERS
---------------
null, Ho: p1 = p2
alternate, H1: p1 != p2
test statistic: 0.972
critical value: -1.96 , 1.96
decision: do not reject Ho
p-value: 0.3311

Minitab output:
Test for difference = 0 (vs ? 0): Z = 0.98 P-Value = 0.328

Fisher’s exact test: P-Value = 0.466

there is evidence that the Penn State faculty and Penn State staff groups are similar in terms of proportion of females ratio

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