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

`4600-1.646(800)/(√(1800))=4568.96`

`4600+1.646(800)/(√(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)

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)(s)/(√(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(800)/(√(1800))=4568.96`

`4600+1.646(800)/(√(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)