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, $\bar x$ = $114.00

Population standard deviation, σ = $15.30

The formula for Confidence Interval, CI is given by the following relation;

$CI=\bar{x}\pm z\frac{\sigma}{\sqrt{n}}$

Where, z is found for the 90% confidence level as ±1.645

Plugging in the values, we have;

$CI=114\pm 1.645 \times \frac{15.3}{\sqrt{45}}$

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 $CI=114\pm 1.96 \times \frac{15.3}{\sqrt{45}}$

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.

8 0

3 years ago

Read 2 more answers

157.70 for 19 shirts

Jlenok [28]

Answer:

Each shirt was $8.30

Step-by-step explanation:

6 0

3 years ago