Question:
You are given the sample mean and the population standard deviation. Use this information to construct the 90% and 95% confidence intervals for the population mean. Interpret the results and compare the widths of the confidence intervals. If convenient, use technology to construct the confidence intervals. A random sample of 45 home theater systems has a mean price of $114.00. Assume the population standard deviation is $15.30. Construct a 90% confidence interval for the population mean.
Answer:
At the 90% confidence level, confidence interval = 110.2484 < μ < 117.7516
At the 95% confidence level, confidence interval = 109.53 < μ < 118.48
The 95% confidence interval is wider
Step-by-step explanation:
Here, we have
Sample size, n = 45
Sample mean, [tex]\bar x[/tex] = $114.00
Population standard deviation, σ = $15.30
The formula for Confidence Interval, CI is given by the following relation;
[tex]CI=\bar{x}\pm z\frac{\sigma}{\sqrt{n}}[/tex]
Where, z is found for the 90% confidence level as ±1.645
Plugging in the values, we have;
[tex]CI=114\pm 1.645 \times \frac{15.3}{\sqrt{45}}[/tex]
or CI: 110.2484 < μ < 117.7516
At 95% confidence level, we have our z value given as z = ±1.96
From which we have [tex]CI=114\pm 1.96 \times \frac{15.3}{\sqrt{45}}[/tex]
Hence CI: 109.53 < μ < 118.48
To find the wider interval, we subtract their minimum from the maximum as follows;
90% Confidence level: 117.7516 - 110.2484 = 7.5
95% Confidence level: 118.47503 - 109.5297 = 8.94
Therefore, the 95% confidence interval is wider.
Answer:
B. is the correct option.
With 90% the population mean price is
in lies in (127.01,134.99). With 95% confidence, it can be said that the population mean price lies in ( 126.24,135.76)
Therefore, the 95% confidence interval is wider than 90%.
The calculation is attached
