Math Problem Statement

A​ country's education department reported that in 2015​, 63.8​% of students enrolled in college or a trade school within 12 months of graduating high school. In 2017​, a random sample of 154 individuals who graduated from high school 12 months prior was selected. From this​ sample, 94 students were found to be enrolled in college or a trade school. Construct a 95​% confidence interval to estimate the actual proportion of students enrolled in college or a trade school within 12 months of graduating from high school in 2017.

The confidence interval has a lower limit of    enter your response here and an upper limit of    enter your response here.

​(Round to three decimal places as​ needed.)

Solution

To construct the 95% confidence interval for the proportion of students enrolled in college or a trade school within 12 months of graduating high school in 2017, follow these steps:

Step 1: Calculate the sample proportion p^\hat{p}

The sample proportion p^\hat{p} is given by:

p^=xn\hat{p} = \frac{x}{n} where:

  • x=94x = 94 (number of students enrolled in college or trade school),
  • n=154n = 154 (total number of students in the sample).

p^=94154=0.610\hat{p} = \frac{94}{154} = 0.610

Step 2: Determine the standard error (SE)

The standard error for the proportion is calculated using the formula:

SE=p^(1p^)nSE = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}

SE=0.610(10.610)154=0.610×0.390154=0.2379154=0.001545=0.0393SE = \sqrt{\frac{0.610(1 - 0.610)}{154}} = \sqrt{\frac{0.610 \times 0.390}{154}} = \sqrt{\frac{0.2379}{154}} = \sqrt{0.001545} = 0.0393

Step 3: Find the critical value for a 95% confidence level

For a 95% confidence level, the critical value zα/2z_{\alpha/2} is 1.96 (from the standard normal distribution table).

Step 4: Calculate the margin of error (ME)

The margin of error is given by:

ME=zα/2×SE=1.96×0.0393=0.077ME = z_{\alpha/2} \times SE = 1.96 \times 0.0393 = 0.077

Step 5: Construct the confidence interval

The confidence interval is calculated as:

p^±ME=0.610±0.077\hat{p} \pm ME = 0.610 \pm 0.077

Thus, the 95% confidence interval is:

(0.6100.077,0.610+0.077)=(0.533,0.687)(0.610 - 0.077, 0.610 + 0.077) = (0.533, 0.687)

Final Answer:

  • The lower limit of the confidence interval is 0.533.
  • The upper limit of the confidence interval is 0.687.

Would you like further details or clarifications on the calculation steps?

Here are 5 related questions to consider:

  1. How would a larger sample size affect the confidence interval?
  2. What would happen to the confidence interval if we used a 99% confidence level?
  3. How is the critical value for the confidence level determined?
  4. How do you interpret the confidence interval in terms of real-world significance?
  5. How does the 2017 sample compare to the 2015 proportion of 63.8%?

Tip: A smaller confidence interval (with a narrower range) indicates more precision, but it requires a larger sample size or lower confidence level.

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Math Problem Analysis

Mathematical Concepts

Statistics
Confidence Intervals
Proportions

Formulas

Sample Proportion (p̂) = x/n
Standard Error (SE) = sqrt[p̂(1 - p̂) / n]
Margin of Error (ME) = z * SE
Confidence Interval = p̂ ± ME

Theorems

Central Limit Theorem

Suitable Grade Level

Undergraduate Statistics or Advanced High School