Question

In: Statistics and Probability

Comparing two population proportions. We expect that there is no difference in proportion of status of...

Comparing two population proportions.

We expect that there is no difference in proportion of status of employment between male and female recent business graduates.

a) insert a frequency table and a bar chart or a pie chart labeled properly. USING EXCEL

b) Perform hypothesis test: Calculate the P-value and make the conclusion (reject or fail to reject Ho). Insert Excel software output.

C)

Calculate the corresponding confidence interval and check if the conclusion is the same

Status	Gender
Part-time	F
Part-time	M
Full-time	M
Full-time	F
Part-time	M
Part-time	M
Part-time	F
Full-time	F
Full-time	F
Full-time	F
Full-time	F
Full-time	F
Part-time	F
Full-time	M
Part-time	F
Full-time	F
Full-time	F
Part-time	M
Part-time	M
Part-time	F
Part-time	F
Part-time	F
Full-time	M
Part-time	F
Part-time	M
Part-time	M
Full-time	M
Part-time	M
Full-time	M
Full-time	F
Part-time	F
Part-time	M
Part-time	M
Part-time	M
Full-time	F
Full-time	M
Part-time	F
Part-time	F
Full-time	F
Full-time	M
Full-time	F
Full-time	F
Full-time	F
Part-time	M
Full-time	M
Full-time	M
Part-time	M
Part-time	M
Full-time	F
Part-time	M
Part-time	F
Full-time	M
Part-time	M
Part-time	M
Full-time	F
Part-time	F
Full-time	M
Part-time	M
Part-time	F
Part-time	M
Part-time	M
Part-time	F
Part-time	F
Part-time	F

Expert Solution

Part a

The required bar chart for the given data by using excel is given as below:

Part b

b) Perform hypothesis test: Calculate the P-value and make the conclusion (reject or fail to reject Ho). Insert Excel software output.

Solution:

Here, we have to use z test for the difference in population proportions.

H₀: p₁ = p₂ versus H₁: p₁ ≠ p₂

This is a two tailed test.

We assume α = 0.05

The test statistic formula is given as below:

Z = (P1 – P2) / sqrt(P*(1 – P)*((1/N1) + (1/N2)))

Where,

X1 = 11

X2 = 16

N1 = 31

N2 = 33

P = (X1+X2)/(N1+N2) = (11+16) / (31+33) = 0.4219

P1 = X1/N1 = 11/31 = 0.35483871

P2 = X2/N2 = 16/33 = 0.484848485

Z = (0.35483871 – 0.484848485) / sqrt(0.4219*(1 - 0.4219)*((1/31) + (1/33)))

Z = -0.13000978 / 0.123526011

Z = -1.0525

Test statistic = Z = -1.0525

P-value = 0.2926

(by using z-table)

P-value > α = 0.05

So, we do not reject the null hypothesis

There is sufficient evidence to conclude that there is no difference in proportion of status of employment between male and female recent business graduates at α = 0.05 level of significance.

Excel output for above test is given as below:

Z Test for Differences in Two Proportions

Data
Hypothesized Difference	0
Level of Significance	0.05
Group 1
Number of Items of Interest	11
Sample Size	31
Group 2
Number of Items of Interest	16
Sample Size	33

Intermediate Calculations
Group 1 Proportion	0.35483871
Group 2 Proportion	0.484848485
Difference in Two Proportions	-0.13000978
Average Proportion	0.4219
Z Test Statistic	-1.0525

Two-Tail Test
Lower Critical Value	-1.9600
Upper Critical Value	1.9600
p-Value	0.2926
Do not reject the null hypothesis

C) Calculate the corresponding confidence interval and check if the conclusion is the same

Confidence interval for difference between two population proportions:

Confidence interval = (P1 – P2) ± Z*sqrt[(P1*(1 – P1)/N1) + (P2*(1 – P2)/N2)]

Where, P1 and P2 are sample proportions for first and second groups respectively.

Confidence level = 95%

Critical Z value = 1.96

(by using z-table)

Confidence interval = (0.35483871 – 0.484848485) ± 1.96*sqrt[(0.35483871*(1 – 0.35483871)/31) + (0.484848485*(1 – 0.484848485)/33)]

Confidence interval = -0.1300 ± 0.2397

Lower limit = -0.1300 - 0.2397 = -0.3697

Upper limit = -0.1300 + 0.2397 = 0.1097

The required confidence interval by using excel is given as below:

Confidence Interval Estimate
of the Difference Between Two Proportions

Data
Confidence Level	95%

Intermediate Calculations
Z Value	1.9600
Std. Error of the Diff. between two Proportions	0.1223
Interval Half Width	0.2397

Confidence Interval
Interval Lower Limit	-0.3697
Interval Upper Limit	0.1097

The 95% confidence interval for the difference in population proportions is given as (-0.3697, 0.1097). The value zero is lies within above confidence interval, so we do not reject the null hypothesis and concluded that there is no difference in proportion of status of employment between male and female recent business graduates.

orchestra answered 1 year ago

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