Questions: In a random sample of four mobile devices, the mean repair cost was 90.00 and the standard deviation was 11.50. Assume the population is normally distributed and use a t-distribution to find the margin of error and construct a 99% confidence interval for the population mean. Interpret the results. The 99% confidence interval for the population mean μ is (56.41,123.59). (Round to two decimal places as needed.) The margin of error is (Round to two decimal places as needed.)

In a random sample of four mobile devices, the mean repair cost was 90.00 and the standard deviation was 11.50. Assume the population is normally distributed and use a t-distribution to find the margin of error and construct a 99% confidence interval for the population mean. Interpret the results. The 99% confidence interval for the population mean μ is (56.41,123.59). (Round to two decimal places as needed.) The margin of error is (Round to two decimal places as needed.)
Transcript text: Part 2 of 3 Points: 0 of 10 Save In a random sample of four mobile devices, the mean repair cost was $\$ 90.00$ and the standard deviation was $\$ 11.50$. Assume the population is normally distributed and use a t-distribution to find the margin of error and construct a $99 \%$ confidence interval for the population mean. Interpret the results. The $99 \%$ confidence interval for the population mean $\mu$ is ( $56.41,123.59$ ). (Round to two decimal places as needed.) The margin of error is $\$$ $\square$ (Round to two decimal places as needed.)
failed

Solution

failed
failed

Solution Steps

Step 1: Calculate the Z-Score

For a \(99\%\) confidence level, the Z-score is given by \(Z = 2.58\).

Step 2: Calculate the Margin of Error

The margin of error (\(E\)) is calculated using the formula:

\[ E = \frac{Z \times \sigma}{\sqrt{n}} \]

Substituting the values:

\[ E = \frac{2.58 \times 11.5}{\sqrt{4}} = \frac{29.67}{2} = 14.81 \]

Thus, the margin of error is \(14.81\).

Step 3: Construct the Confidence Interval

The \(99\%\) confidence interval for the population mean (\(\mu\)) is calculated as follows:

\[ \text{Lower Bound} = \text{Mean} - E = 90.00 - 14.81 = 75.19 \]

\[ \text{Upper Bound} = \text{Mean} + E = 90.00 + 14.81 = 104.81 \]

Therefore, the \(99\%\) confidence interval is \((75.19, 104.81)\).

Final Answer

The margin of error is \(\boxed{14.81}\) and the \(99\%\) confidence interval for the population mean \(\mu\) is \(\boxed{(75.19, 104.81)}\).

Was this solution helpful?
failed
Unhelpful
failed
Helpful

Questions asked by other users

failedAnswered
About what percent of x values lie between the first and third standard deviations (both sides)? (Enter an exact number as an integer, fraction, or decimal.)
failedAnswered
Determine the acid-base disorder present: Interpret the following ABG: pH 7.21 PaCO2 33 HCO 12
failedAnswered
The degree of f(x) is □ and the leading coefficient is □ . There are □ distinct real zeros and □ relative maximums.
failedAnswered
After for traveling for 2 days, Adriel has visited 12 states. What is Adriel's rate?
failedAnswered
Which best illustrates a small business operating in a retail industry as opposed to wholesaling its products? Multiple Choice Moriah bakes doughnuts and sells them to restaurants. Sticking It manufactures unique stickers and sells them in bulk to office supply stores. Iron Heat, an apparel company, sells its merchandise to clothing retailers. Perfect Suds imports soap and directly sells it to customers through kiosks. Chip To It manufactures computer parts that are resold by tech companies.