Question

The high price of medicines is a source of major expense for those seniors in the United States who have to pay for these medicines themselves. A random sample of 1800 seniors who pay for their medicines showed that they spent an average of $ 4600 last year on medicines with a standard deviation of $ 800 . Make a 90 % confidence interval for the corresponding population mean.

Answer 1

Answer:

$4600-1.646\frac{800}{\sqrt{1800}}=4568.96$    

$4600+1.646\frac{800}{\sqrt{1800}}=4631.04$    

We are confident at 90% that the true mean for the amount spent on medicines is between (4568.96 and 4631.04)

Step-by-step explanation:

Information given

$\bar X=4600$ represent the sample mean for the amount spent on medicines

$\mu$ population mean

s=800 represent the sample standard deviation

n=1800 represent the sample size  

Solution

The confidence interval for the true population mean is given by:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$   (1)

We can calculate the degrees of freedom with:

$df=n-1=1800-1=1799$

We know that the Confidence level is 0.90 or 90%, the value of significance is $\alpha=0.1$ 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(0.05,1799)".And we see that $t_{\alpha/2}=1.646$

Replcing in the formula for the confidence interval we got:

$4600-1.646\frac{800}{\sqrt{1800}}=4568.96$    

$4600+1.646\frac{800}{\sqrt{1800}}=4631.04$    

We are confident at 90% that the true mean for the amount spent on medicines is between (4568.96 and 4631.04)