Answer:
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!!!
