UkusakazaL
2021-08-10
Answered

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 2000 seniors who pay for their medicines showed that they spent an average of $\mathrm{\$}4600}$ last year on medicines with a standard deviation of $\mathrm{\$}800}$ .
a) Make a $98\mathrm{\%}$ confidence interval for the corresponding population mean.
b) Suppose the confidence interval obtained in part a is too wide. How can the width of this interval be reduced? Discuss allpossible alternatives. Which alternative is the best?

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fortdefruitI

Answered 2021-07-30
Author has **1** answers

a) First note that it is a one sample test and the population standard deviation is not given. Hence t-distribution will be appropriate in this scenario.

Sample mean,$\stackrel{\u2015}{x}=4600$

Sample standard deviation$s=800$

Number of samples,$n=2000$

Hence,$100(1-\alpha )\mathrm{\%}$ Confidence interval is given by,

$(\stackrel{\u2015}{x}-\frac{s}{\sqrt{n}}{t}_{n-1,\text{}\frac{\alpha}{2}},\text{}\stackrel{\u2015}{x}+\frac{s}{\sqrt{n}}{t}_{n-1,\text{}\frac{\alpha}{2}})$

Here,$\alpha =0.01$

Hence, Confidence interval is given by,

$(4558.352,\text{}4641.648)$

b) Now from the equation of confidence interval,$(\stackrel{\u2015}{x}-\frac{s}{\sqrt{n}}{t}_{n-1,\text{}\frac{\alpha}{2}},\text{}\stackrel{\u2015}{x}+\frac{s}{\sqrt{n}}{t}_{n-1,\text{}\frac{\alpha}{2}})$ we can see that the range will decrease if n increases or the t value decreases. t value will decrease when we decreases $\alpha$ keeping the n as fixed. Or increase n when α is fixed. The best way to decrease range is to increase the sample size.

Sample mean,

Sample standard deviation

Number of samples,

Hence,

Here,

Hence, Confidence interval is given by,

b) Now from the equation of confidence interval,

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