You are given the sample mean and the population standard deviation. Use this information to construct the​ 90% and​ 95% confidence interval

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. Which interval is​ wider? If​ convenient, use technology to construct the confidence intervals.

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Delilah 2 days 2021-10-13T07:15:44+00:00 2 Answers 0

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    0
    2021-10-13T07:16:44+00:00

    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

    0
    2021-10-13T07:17:02+00:00

    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, \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.

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