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Use the t-distribution to find a confidence interval for a mean μ given the relevant sample results. Give the best point estimate for μ, the margin of error, and the confidence interval. Assume the results come from a random sample from a population that is approximately normally distributed.1. A 90% confidence interval for μ using the sample results x^- =143.0, s=56.7, and n=50Round your answer for the point estimate to one decimal place, and your answers for the margin of error and the confidence interval to two decimal places.i. point estimate = ii. margin of error = iii. The 90% confidence interval is_______ to _________.

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Answer:

i. Point of estimate:


\hat \mu = \bar X =143.0

ii. Margin of error:


ME = 2.01 *(56.7)/(√(50))= 16.12

iii. The 90% confidence interval

Replacing in the confidence interval formula we got


143.0-16.12=126.88


143.0+16.12=159.12

The 90% confidence interval is 126.88 to 159.12

Explanation:

Information given


\bar X=143.0 represent the sample mean for the variable of interest


\mu population mean

s=56.7 represent the sample standard deviation

n=50 represent the sample size

Confidence interval

The confidence interval for the true mean when we don't know the deviation is given by the following formula:


\bar X \pm t_(\alpha/2)(s)/(√(n)) (1)

In order to calculate the critical value for the confidence interval
t_(\alpha/2) we need to find the degrees of freedom, with this formula:


df=n-1=50-1=49

The Confidence level provided is 0.90 or 90%, the value for the significance is
\alpha=1-0.9=0.1 and
\alpha/2 =0.05, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.05,49)".And we see that
t_(\alpha/2)=2.01

i. Point of estimate:


\hat \mu = \bar X =143.0

ii. Margin of error:


ME = 2.01 *(56.7)/(√(50))= 16.12

iii. The 90% confidence interval

Replacing in the confidence interval formula we got


143.0-16.12=126.88


143.0+16.12=159.12

The 90% confidence interval is 126.88 to 159.12

User Thomas Weglinski
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