Question

question 3 of 10
a 95% confidence interval is found to be (30, 36). the sample standard deviation is 9, and the sample size is 36. if you wanted a smaller interval, what could you do?
a. none of these
b. increase the confidence level to 99.7%.
c. increase the sample size to 100.
d. decrease the sample size to 30.

Explanation:

Step1: Recall the formula for confidence interval width

The width of a confidence interval for a population mean (when population standard deviation \(\sigma\) is unknown and we use sample standard deviation \(s\)) is \(w = 2\times t_{\alpha/2}\times\frac{s}{\sqrt{n}}\). Here, \(s = 9\). For a large - enough \(n\) (by Central Limit Theorem, when \(n\geq30\), we can approximate \(t\) - distribution with \(z\) - distribution). The formula for the margin of error \(E=z_{\alpha/2}\times\frac{s}{\sqrt{n}}\) (width \(w = 2E\)).

Step2: Analyze each option

• Option B:

As the confidence level increases, the value of \(z_{\alpha/2}\) increases. For a 95% confidence interval, \(z_{\alpha/2}=1.96\). For a 99.7% confidence interval (\(\alpha=1 - 0.997=0.003\), \(\alpha/2 = 0.0015\)), \(z_{\alpha/2}=3\). Using the formula \(E = z_{\alpha/2}\times\frac{s}{\sqrt{n}}\), when \(z_{\alpha/2}\) increases (while \(s\) and \(n\) are fixed), the margin of error \(E\) (and thus the width of the confidence interval) increases.

• Option C:

Given \(s = 9\), originally \(n = 36\), \(E_1=z_{\alpha/2}\times\frac{s}{\sqrt{n_1}}\). If \(n_2 = 100\), then \(E_2=z_{\alpha/2}\times\frac{s}{\sqrt{n_2}}\). Substitute \(s = 9\), \(n_1 = 36\), \(n_2 = 100\) into the formula. \(E_1=z_{\alpha/2}\times\frac{9}{\sqrt{36}}=z_{\alpha/2}\times1.5\), \(E_2=z_{\alpha/2}\times\frac{9}{\sqrt{100}}=z_{\alpha/2}\times0.9\). Since \(E_20\)), increasing the sample size will decrease the margin of error (and thus the width of the confidence interval).

• Option D:

If \(n\) decreases from \(n_1 = 36\) to \(n_3 = 30\), using the formula \(E=z_{\alpha/2}\times\frac{s}{\sqrt{n}}\), since \(\frac{1}{\sqrt{n}}\) is a decreasing function of \(n\) for \(n>0\), \(E\) (and thus the width of the confidence interval) will increase.

Answer:

C. Increase the sample size to 100.