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% confid

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?

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Aaliyah 3 days 2021-10-10T23:55:41+00:00 1 Answer 0

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    2021-10-10T23:56:51+00:00

    Answer:

    When confidence level increases, margin of error increases thus making confidence interval wider.

    Step-by-step explanation:

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

    Since n >30 but population std deviation is not known we can use only t critical value.

    a) t critical = 2.006

    Hence 90% confidence interval = Mean ±2.006*\frac{4.4}{\sqrt{54} }

    =(22.5-1.201, 22.5+1.201)\\\\= (21.299,23.701)

    b) t critical = 2.30687 = 2.31

    Hence 90% confidence interval = Mean ±2.31*\frac{4.4}{\sqrt{54} }

    =(22.5-1.3813, 22.5+1.3814)\\\\= (21.117,23.883)

    c) t critical = 2.30687 = 2.31

    Hence 90% confidence interval = Mean ±2.31*\frac{4.4}{\sqrt{54} }

    =(22.5-1.754, 22.5+1.754)\\\\= (20.746,24.254)

    d) When confidence level increases, margin of error increases thus making confidence interval wider.

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