A simple random sample with provided a sample mean of and a sample standard deviation of . a. Develop a confidence interval for the populati

Question

A simple random sample with provided a sample mean of and a sample standard deviation of . a. Develop a confidence interval for the population mean (to 1 decimal). , b. Develop a confidence interval for the population mean (to 1 decimal). , c. Develop a confidence interval for the population mean (to 1 decimal). , d. What happens to the margin of error and the confidence interval as the confidence level is increased?

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Raelynn 2 weeks 2021-11-19T18:44:47+00:00 1 Answer 0 views 0

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    2021-11-19T18:45:47+00:00

    Question:

    A simple random sample with n=54 provided a sample mean of 22.5 and a sample standard deviation of 4.4.

    a. Develop a 90%  confidence interval for the population mean.

    b. Develop a 95% confidence interval for the population mean.

    c. Develop a 99%  confidence interval for the population mean.

    d. What happens to the margin of error and the confidence interval as the confidence level is increased?

    Answer:

    a. CI = 21.4976 ≤ μ ≤ 23.5024

    b. CI = 21.29903 ≤ μ ≤ 23.70097

    c. CI = 20.90021 ≤ μ ≤ 24.09979

    d. The margin of error and confidence interval increases

    Step-by-step explanation:

    Here we have

    Sample size, n = 54

    Mean, \bar{x} = 22.5

    s = 4.4

    a. The confidence interval, CI is;

    CI=\bar{x}\pm t_{\alpha/2} \times \frac{s}{\sqrt{n}}

    At c = 90% confidence level, t_{\alpha/2} with degrees of freedom, df = n – 1 = 54 – 1 = 53 and α = (1 – c)/2 = (1 – 0.9)/2 = 0.05, from the t table or relation we will find

    t_{\alpha/2}  = 1.674116

    Therefore, plugging in the values, we have

    CI at 90% gives

    CI=22.5\pm 1.674 \times \frac{4.4}{\sqrt{54}}

    or CI = 21.4976 ≤ μ ≤ 23.5024

    b. For c = 95%, α = (1 – 0.95)/2 = 0.025, df = 54 – 1 = 53

    From tables or t relations, t_{\alpha/2}  = 2.005746

    CI=22.5\pm 2.005746 \times \frac{4.4}{\sqrt{54}}

    or CI = 21.29903 ≤ μ ≤ 23.70097

    c. For c = 99%, α = (1 – 0.99)/2 = 0.005, df = 54 – 1 = 53

    From tables or t relations, t_{\alpha/2}  = 2.671822

    CI=22.5\pm 2.671822\times \frac{4.4}{\sqrt{54}}

    or CI = 20.90021 ≤ μ ≤ 24.09979

    d. As the confidence level is increased, t_{\alpha/2}   increases, therefore, the margin of error and confidence interval increases.

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