Question

Data concerning employment status were collected from a sample of 100 World Campus students at the beginning of the Fall 2018 semester. In that sample of 100, 67 were employed full-time.

1. We want use these data to construct a 95% confidence interval to estimate the proportion of all World Campus students who are employed full-time. Is it appropriate to use the normal approximation method? Show all your work involved with checking these assumptions.
2. 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 in part a, use the normal approximation method. Remember to include all relevant Minitab Express output and to clearly identify your answer. Do not do any hand calculations.
3. 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 5%? You will need to do hand calculations. Show all your work for full credit with correct answers.
4. 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 Minitab Express and the five-step hypothesis testing procedure given below. Remember to include all relevant Minitab Express output. Do not do any hand calculations.  

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.

Answer 1

(a) Since np = 67 and n(1 - p) = 33 are both > 5, normal approximation can be used to construct the confidence interval

(b)

est and CI for Oe Proportion Method p: event proportion Normal approximation method is used for this analysis Descriptive Statistics N Event Sample p 95% CI for p 100 67 0.670000 (0.577840, 0.762160)

The 95% confidence interval is [0.5778, 0.7622]

(c) z- score for 95% confidence is z = 1.96

N = (z/E)^2 * p(1 - p) = (1.96/0.05)^2 * 0.67 * 0.33 = 340

(d)

Hypotheses, Ho: p ≤ 0.6 versus Ha: p > 0.6

Test and Cl for One Proportion Method p: event proportion Normal approximation method is used for this analysis Descriptive Statistics 95% Lower Bound for p 0.592657 N Event Sample p 100 67 0.670000 Test Null hypothesis Alternative hypothesis H:p 0.6 Z-Value P-Value Ho: p = 0.6 1.43 0.077

Since the p- value > 0.05, we fail to reject Ho

Conclusion: There is no sufficient evidence that more than 60% are employed full-time