Question

a random sample of the closing stock prices in dollars for a company in a recent year is listed below. assume that σ is $1.94. construct the 90% and 99% confidence intervals for the population mean. interpret the results and compare the widths of the confidence intervals.
22.72 18.81 18.18 21.35 18.15 20.96 22.42 19.55
21.76 19.64 21.59 22.39 15.49 19.16 18.98 19.09
the 90% confidence interval is ($□,$□)
(round to two decimal places as needed.)
the 99% confidence interval is ($□,$□)
(round to two decimal places as needed.)
which statement below interprets the results correctly?
the probability that the mean closing stock price is in the 90% confidence interval is about 90% and the probability that the mean closing stock price is in the 99% confidence interval is about 99%.
90% of the mean closing stock prices are in the 90% confidence interval and 99% of the mean closing stock prices are in the 99% confidence interval
the 90% confidence interval contains the mean closing stock price 90% of the time and the 99% confidence interval contains the mean closing stock price 99% of the time
there is 90% confidence that the mean closing stock price is in the 90% confidence interval and 99% confidence that the mean closing stock price is in the 99% confidence interval.
which interval is wider?
the 99% confidence interval
the 90% confidence interval

Explanation:

Step1: Calculate the sample mean

First, find the sum of the data values:

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The sample size \(n = 16\). The sample mean \(\bar{x}=\frac{337.24}{16}=21.08\)

Step2: Find the z - values

For a \(90\%\) confidence interval, the significance level \(\alpha=1 - 0.90=0.10\), and \(\frac{\alpha}{2}=0.05\). The \(z\) - value \(z_{\frac{\alpha}{2}}=z_{0.05}\approx1.645\) (from the standard normal distribution table).
For a \(99\%\) confidence interval, the significance level \(\alpha = 1-0.99 = 0.01\), and \(\frac{\alpha}{2}=0.005\). The \(z\) - value \(z_{\frac{\alpha}{2}}=z_{0.005}\approx2.576\)

Step3: Calculate the margin of error for \(90\%\) confidence interval

The formula for the margin of error \(E = z_{\frac{\alpha}{2}}\frac{\sigma}{\sqrt{n}}\). Given \(\sigma = 1.94\), \(n = 16\)
\(E_{90}=1.645\times\frac{1.94}{\sqrt{16}}=1.645\times\frac{1.94}{4}=1.645\times0.485 = 0.80\)
The \(90\%\) confidence interval is \(\bar{x}-E_{90}<\mu<\bar{x} + E_{90}\), \(21.08- 0.80<\mu<21.08 + 0.80\), \(20.28<\mu<21.88\)

Step4: Calculate the margin of error for \(99\%\) confidence interval

\(E_{99}=2.576\times\frac{1.94}{\sqrt{16}}=2.576\times\frac{1.94}{4}=2.576\times0.485=1.25\)
The \(99\%\) confidence interval is \(\bar{x}-E_{99}<\mu<\bar{x}+E_{99}\), \(21.08-1.25 <\mu<21.08 + 1.25\), \(19.83<\mu<22.33\)

Step5: Interpret the confidence intervals

The interpretation of a confidence interval: There is \(C\%\) confidence that the population mean \(\mu\) lies within the interval. So, there is \(90\%\) confidence that the mean closing stock price is in the \(90\%\) confidence interval and \(99\%\) confidence that the mean closing stock price is in the \(99\%\) confidence interval.
Since \(E_{99}>E_{90}\), the \(99\%\) confidence interval is wider.

Answer:

The \(90\%\) confidence interval is \((20.28,21.88)\)
The \(99\%\) confidence interval is \((19.83,22.33)\)
The correct interpretation: There is \(90\%\) confidence that the mean closing stock price is in the \(90\%\) confidence interval and \(99\%\) confidence that the mean closing stock price is in the \(99\%\) confidence interval.
The wider interval: The \(99\%\) confidence interval.