Question

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 _________.

Answer 1

Answer:

i. Point of estimate:

$\hat \mu = \bar X =143.0$

ii. Margin of error:

$ME = 2.01 *\frac{56.7}{\sqrt{50}}= 16.12$

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

Step-by-step 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:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$   (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:

$ME = 2.01 *\frac{56.7}{\sqrt{50}}= 16.12$

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