Question

2. Data concerning employment status were collected from a sample of 50 World Campus students. In that sample of 50 students, 33 students reported they were employed full-time.

A. Use Minitab Express to construct a 95% confidence interval to estimate the proportion of all World Campus students who are employed full-time. If assumptions were met, use the normal approximation method. Remember to include all relevant Minitab Express output and to clearly identify your answer. [15 points]

B. What sample size would be necessary to construct a 95% confidence interval to estimate the proportion of all World Campus students who are employed full-time with a margin of error of 2%? You will need to do hand calculations. Show all of your work. [10 points]

C. We want to know if there is evidence that in the population of all World Campus students, more than 60% are employed full-time. Use the five-step hypothesis testing procedure to address this research question. The only hand calculations that you will need to do will be in step 1 to check assumptions; use Minitab Express for steps 2 and 3. Remember to include all relevant Minitab Express output and to clearly identify your test statistic and p value. [25 points]

Step 1: Check assumptions and write hypotheses

Step 2: Calculate the test statistic

Step 3: Determine the p value

Step 4: Decide to reject or fail to reject the null

Step 5: State a “real world” conclusion

Answer 1

Part a

Here, we have to find the 95% confidence interval for the population proportion.

We are given

x = 33

n = 50

All assumptions are met, because sample size is greater than 30 and it is adequate to use normal approximation.

The required Minitab output is given as below:

Test and CI for One Proportion

Test of p = 0.5 vs p not = 0.5

                                                        Exact

Sample      X      N Sample p         95.0% CI       P-Value

1          33     50 0.660000 (0.512348, 0.787945)    0.033

Confidence interval = (0.5123, 0.7879)

We are 95% confident that the population proportion of all world campus students who are employed full time will lies within 0.5123 and 0.7879.

Part b

We are given

Confidence level = 95%

Z = 1.96

Margin of error = E = 2% = 0.02

We are given

x = 33

n = 50

Estimate for proportion = p = x/n = 33/50 = 0.66

q = 1 – p = 1 – 0.66 = 0.34

Sample size formula is given as below:

n = p*q*(Z/E)^2

n = 0.66*0.34*(1.96/0.02)^2

n = 2155.138

n = 2156

Part c

Here, we have to use z test for population proportion.

H₀: p = 0.60 versus H_a: p > 0.60

This is an upper tailed test.

We assume 5% level of significance.

α = 0.05

The test statistic formula for this test is given as below:

Z = (p̂ - p)/sqrt(pq/n)

Where, p̂ = Sample proportion, p is population proportion, q = 1 - p, and n is sample size

p̂ = x/n

x = number of items of interest

We are given

x = 33

n = 50

p̂ = x/n = 33/50 = 0.66

Z = (0.66 – 0.60)/sqrt(0.60*0.40/50)

Z = 0.8660

P-value = 0.1932

(by using z-table)

P-value > α = 0.05

So, we do not reject the null hypothesis

There is insufficient evidence to conclude that the population of all World Campus students who employed full time is more than 60%.