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.)

A) 95% confidence interval estimate for the proportion of engineers remaining in the profession = (0.486, 0.614)

Lower limit = 0.486

Upper limit = 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 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!!!