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 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.

Confidence intervals
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asked 2021-08-10
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 \(\displaystyle\${4600}\) last year on medicines with a standard deviation of \(\displaystyle\${800}\). a) Make a \(\displaystyle{98}\%\) 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?

Expert Answers (1)

2021-07-30
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, \(\displaystyle\overline{{{x}}}={4600}\)
Sample standard deviation \(\displaystyle{s}={800}\)
Number of samples, \(\displaystyle{n}={2000}\)
Hence, \(\displaystyle{100}{\left({1}-\alpha\right)}\%\) Confidence interval is given by,
\(\displaystyle{\left(\overline{{{x}}}-{\frac{{{s}}}{{\sqrt{{{n}}}}}}{t}_{{{n}-{1},\ \frac{\alpha}{{2}}}},\ \overline{{{x}}}+{\frac{{{s}}}{{\sqrt{{{n}}}}}}{t}_{{{n}-{1},\ \frac{\alpha}{{2}}}}\right)}\)
Here, \(\displaystyle\alpha={0.01}\)
Hence, Confidence interval is given by,
\(\displaystyle{\left({4558.352},\ {4641.648}\right)}\)
b) Now from the equation of confidence interval, \(\displaystyle{\left(\overline{{{x}}}-{\frac{{{s}}}{{\sqrt{{{n}}}}}}{t}_{{{n}-{1},\ \frac{\alpha}{{2}}}},\ \overline{{{x}}}+{\frac{{{s}}}{{\sqrt{{{n}}}}}}{t}_{{{n}-{1},\ \frac{\alpha}{{2}}}}\right)}\) we can see that the range will decrease if n increases or the t value decreases. t value will decrease when we decreases \(\displaystyle\alpha\) 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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