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.

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 $4600 last year on medicines with a standard deviation of $800. a) Make a 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?
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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, x=4600
Sample standard deviation s=800
Number of samples, n=2000
Hence, 100(1α)% Confidence interval is given by,
(xsntn1, α2, x+sntn1, α2)
Here, α=0.01
Hence, Confidence interval is given by,
(4558.352, 4641.648)
b) Now from the equation of confidence interval, (xsntn1, α2, x+sntn1, α2) we can see that the range will decrease if n increases or the t value decreases. t value will decrease when we decreases α 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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Answered 2021-11-03
Loop or thiazide diuretics) hypersensitivity reaction side the doctor's your you found that had higher risk scores, more co-morbidities, and/or used more $4 drugs
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