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Anarel [89]
3 years ago
15

The high price of medicines is a source of major expense for those seniors in the United States who have to pay for these medici

nes 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.
Mathematics
1 answer:
Ilia_Sergeevich [38]3 years ago
6 0

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)

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