Math Problem Statement

A survey of 300 union members in New York State reveals that 112 favor the Republican candidate for governor. Construct a 90% confidence interval for the percentage of all New York State union members who favor the Republican candidate.

Question 8 options:

(30.1%, 44.5%)

(32.7%, 41.9%)

(31.9%, 42.8%)

(26.7%, 47.9%)

(17.8%, 56.8%)

Solution

To construct a 90% confidence interval for the percentage of New York State union members who favor the Republican candidate, we will use the formula for the confidence interval of a proportion.

Step-by-Step Solution:

  1. Sample Proportion (p^\hat{p}):

    • The sample size (nn) is 300.
    • The number of union members who favor the Republican candidate is 112.

    The sample proportion p^\hat{p} is: p^=112300=0.3733\hat{p} = \frac{112}{300} = 0.3733

  2. Critical Value for 90% Confidence Level: The critical value (zα/2z_{\alpha/2}) for a 90% confidence interval can be found from the standard normal distribution. For a 90% confidence level, zα/21.645z_{\alpha/2} \approx 1.645.

  3. Standard Error (SE) of the Proportion: The standard error is given by: SE=p^(1p^)n=0.3733(10.3733)300=0.3733×0.6267300=0.23403000.0279SE = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} = \sqrt{\frac{0.3733(1 - 0.3733)}{300}} = \sqrt{\frac{0.3733 \times 0.6267}{300}} = \sqrt{\frac{0.2340}{300}} \approx 0.0279

  4. Confidence Interval Formula: The confidence interval is calculated as: p^±zα/2×SE\hat{p} \pm z_{\alpha/2} \times SE

    Substituting the values: 0.3733±1.645×0.02790.3733 \pm 1.645 \times 0.0279 0.3733±0.04590.3733 \pm 0.0459

  5. Resulting Confidence Interval: 0.3733 - 0.0459 = 0.3274 \quad \text{(32.7%)} 0.3733 + 0.0459 = 0.4192 \quad \text{(41.9%)}

Therefore, the 90% confidence interval for the percentage of New York State union members who favor the Republican candidate is (32.7%, 41.9%).

The correct option is:

(32.7%, 41.9%)


Would you like more details on any part of the calculation or the reasoning behind confidence intervals?

Here are 5 related questions to consider:

  1. How would the confidence interval change if the sample size were larger?
  2. What happens to the confidence interval if the confidence level is increased to 95%?
  3. Why is the normal distribution used to approximate proportions in large samples?
  4. How does the margin of error affect the width of the confidence interval?
  5. What is the significance of a confidence interval in statistical decision-making?

Tip: A larger sample size decreases the standard error, leading to a narrower confidence interval.

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Math Problem Analysis

Mathematical Concepts

Statistics
Confidence Intervals
Proportions

Formulas

Sample Proportion: p̂ = x/n
Standard Error of a Proportion: SE = sqrt(p̂(1 - p̂) / n)
Confidence Interval: p̂ ± z * SE

Theorems

Central Limit Theorem

Suitable Grade Level

Grades 10-12