11-32 (modified). A statistician is testing the null hypothesis that exactly half of all engineers will still be in the profession 10 years after receiving their bachelor's. She took a random sample of 200 graduates from the class of 1979 and determined their occupations in 1989. She found that 111 persons were still employed primarily as engineers. (a) Construct a 95% confidence interval estimate for the proportion of engineers remaining in the profession. (Ignore the continuity correction and use the formulas on p. 314.)

B) The confidence interval obtained agrees with the null hypothesis and the claim that exactly half of the engineers that graduated with a bachelor's degree are still practicing 10 years after obtaining the degree as the proportion of the claim (0.50) lies in the obtained confidence interval.

Accept H₀.

Also, the p-value for this hypothesis test is greater than the significance level at which the test was performed, hence, we fail to reject the null hypothesis. We accept H₀.

Step-by-step explanation:

A) Confidence Interval for the population proportion is basically an interval of range of values where the true population proportion can be found with a certain level of confidence.

Mathematically,

Confidence Interval = (Sample proportion) ± (Margin of error)

Sample proportion = (111/200) = 0.555

Margin of Error is the width of the confidence interval about the mean.

It is given mathematically as,

Margin of Error = (Critical value) × (standard Error)

Critical value will be obtained using the t-distribution. This is because there is no information provided for the population mean and standard deviation.

To find the critical value from the t-tables, we first find the degree of freedom and the significance level.

Degree of freedom = df = n - 1 = 200 - 1 = 199

Significance level for 95% confidence interval

(100% - 95%)/2 = 2.5% = 0.025

t (0.025, 199) = 1.97 (from the t-tables)

Standard error of the sample proportion = σₓ = √[p(1-p)/n]

p = 0.555

n = sample size = 200

σₓ = √[0.555×0.445/200] = 0.03514

95% Confidence Interval = (Sample proportion) ± [(Critical value) × (standard Error)]

CI = 0.555 ± (1.97 × 0.03514)

CI = 0.555 ± 0.06923

95% CI = (0.48577, 0.61423)

95% Confidence interval = (0.486, 0.614)

B) The confidence interval obtained agrees with the null hypothesis and the claim that exactly half of the engineers that graduated with a bachelor's degree are still practicing 10 years after obtaining the degree as the proportion of the claim (0.50) lies in the obtained confidence interval.

Accept the null hypothesis.

We could go a step further and obtain the p-value at this significance level to be totally sure.

We compute the test statistic

t = (x - μ₀)/σₓ

x = p = sample proportion = 0.555

μ₀ = p₀ = the proportion in the claim = 0.50

σₓ = standard error = 0.03514 (already calculated in (a) above)

t = (0.555 - 0.50) ÷ 0.03514

t = 1.57

checking the tables for the p-value of this t-statistic

Degree of freedom = df = n - 1 = 200 - 1 = 199

Significance level = 0.05

The hypothesis test uses a two-tailed condition because we're testing in both directions.

p-value (for t = 1.57, at 0.05 significance level, df = 199, with a one tailed condition) = 0.118004

The interpretation of p-values is that

When the (p-value > significance level), we fail to reject the null hypothesis and when the (p-value < significance level), we reject the null hypothesis and accept the alternative hypothesis.

So, for this question, significance level = 0.05

p-value = 0.118004

0.118004 > 0.05

Hence,

p-value > significance level

This means that we fail to reject the null hypothesis. We accept H₀.

Hope this Helps!!!