Question

A school administrator wants to know what proportion of the approximately 80,000 teachers in their state have a Master's degree. The administrator takes an SRS of 100 teachers from a statewide database containing every teacher, and they find 55 teachers in the sample have a Master's degree.

Based on this sample, which of the following is a 99% confidence interval for the proportion of teachers in the state with a Master's degree?

A 0.55±0.128
B 0.55±0.098
C 55±0.128
D 55±0.098

Answer 1

The 99% confidence interval for the proportion of teachers in the state with a Master's degree, based on a sample of 100, where 55 have a Master's degree, is 0.55 ± 0.128. The correct answer is option A 0.55±0.128

Step-by-step explanation:

The question asks us to find a 99% confidence interval for the proportion of teachers in the state with a Master's degree based on a sample of 100 teachers where 55 have a Master's degree. The sample proportion (p-hat) is 55/100 = 0.55. To construct a 99% confidence interval, we need to use the formula ± z*sqrt([p-hat(1 - p-hat)]/n), where z* is the z-score corresponding to the desired confidence level and n is the sample size.

Since we're computing a 99% confidence interval, we use a z-score of approximately 2.576 (which we would find in a z-table). Plugging in the values, we get the standard error: 2.576*sqrt([0.55*(1 - 0.55)]/100) ≈ 2.576*sqrt(0.002475) ≈ 0.128.

Therefore, the 99% confidence interval would be 0.55 ± 0.128, which represents option A. Keep in mind that when presenting a proportion, it is accurate to use the decimal form rather than the percentage. Hence, the confidence interval of '55 ± 0.128' is inappropriate as it does not conform to the standard convention for representing proportions.