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.

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

represent the sample mean for the amount spent on medicines

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:

  (1)

We can calculate the degrees of freedom with:

We know that the Confidence level is 0.90 or 90%, the value of significance is and , 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

Replcing in the formula for the confidence interval we got:

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

repeat the question

step-by-step explanation:

for the first question it would be the second answer for the second question it would be the third answer and for the third question it would be the first answer

step-by-step explanation: