# 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 $4600 last year on medicines with a standard deviation of$800. Make a 98\% confidence interval for the corresponding population mean.

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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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, $\stackrel{―}{x}=4600$
Sample standard deviation $s=800$
Number of samples, $n=2000$
Hence, $100\left(1-\alpha \right)\mathrm{%}$ Confidence interval is given by,

Here, $\alpha =0.01$
Hence, Confidence interval is given by,

b) Now from the equation of confidence interval, 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.
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