Amalie wants to know if students in her class prefer hot dogs or hamburgers. Why should Amalie survey the entire population rath
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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, = $114.00
Population standard deviation, σ = $15.30
The formula for Confidence Interval, CI is given by the following relation;
Where, z is found for the 90% confidence level as ±1.645
Plugging in the values, we have;
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
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.