Question

Harini is an undergraduate student looking for a summer internship in Statistics or a related field. She observes that the proportion of undergraduate internship postings for all four fields in 2016 was 35.8%. Using this as a planning value, she wants to estimate the proportion for the upcoming summer at a 90% confidence level with a margin of error no greater than 3.47%.

How many internship postings would Harini need to sample to achieve this goal? Please provide the number of internship postings she needs to sample.

After discussing her interests with professors and older students, Harini decides on government internships. She takes a sample of 215 government internship postings for the upcoming summer and finds that 32 are offered to undergraduate students. Construct a 95% confidence interval for the true proportion of internship postings in government that are offered to undergraduates.

Harini's older brother, Harish, is interested in the proportion of internships offered to graduate students across all four fields. He collects a random sample of 225 internship postings and constructs a 95% confidence interval for the true proportion, resulting in (0.4987, 0.6313).

Which of the following statements are appropriate interpretations of this result? Select all that apply:
A. There is a 95% chance that the true proportion of internship postings for graduate students is between 0.4987 and 0.6313.
B. 95% of the time, Harish's calculated confidence interval will be accurate.
C. If Harish took 100 random samples of the same size and constructed 100 confidence intervals, he could expect approximately 95 of the intervals to contain the true proportion of internship postings for graduate students.
D. Harish can be 95% confident that the true proportion of internship postings for graduate students is between 0.4987 and 0.6313.

If Harish decides to calculate a 90% confidence interval for the true proportion of internship postings offered to graduate students instead of a 95% interval, how will the new interval compare to the 95% interval? (Select one of the options: "Wider," "Narrower," or "About the same.")

Answer 1

Harini would need to sample 736 internship postings to achieve her goal of estimating the proportion for the upcoming summer with a 90% confidence level and a margin of error no greater than 3.47%.

For the 95% confidence interval for the true proportion of government internship postings offered to undergraduates, the interval is approximately
`\(0.1180\)` to
`\(0.2256\)`.

Regarding Harish's 90% confidence interval for the true proportion of internship postings offered to graduate students, it will be narrower than the 95% interval.

Explanation:

To calculate the required sample size for Harini's goal, she uses the formula for sample size determination based on the planning value:

`\[ n = \left(\frac{Z_(\alpha/2) \cdot \sqrt{\hat{p}(1-\hat{p})}}{E}\right)^2 \]`

Given the planning value
`\(\hat{p} = 0.358\)`, desired margin of error
`\(E = 0.0347\)`, and confidence level of
`\(90%\)`, she calculates the sample size
`\(n\)`. The value of
`\(Z_(\alpha/2)\)` for a 90% confidence level is approximately
`\(1.645\)`. Plugging these values into the formula yields
`\(n \approx 736\)`.

For the 95% confidence interval of government internship postings offered to undergraduates, Harini uses the sample proportion
`\( \hat{p} = (x)/(n) = (32)/(215) \approx 0.1488 \)` and the standard error formula to calculate the interval. The interval is computed as approximately
`\(0.1480\) to \(0.2256\)`.

Harish's 95% confidence interval for graduate student internship postings is
`\( (0.4987, 0.6313) \)`. Option C best interprets the result: If Harish took 100 random samples of the same size and constructed 100 confidence intervals, approximately 95 of those intervals would contain the true proportion of internship postings for graduate students.

If Harish calculates a 90% confidence interval instead of a 95% interval, the new interval will be narrower than the 95% interval. A higher confidence level requires wider intervals to accommodate a larger margin of error, so decreasing the confidence level to 90% will result in a narrower interval for the same sample size and proportion.