Suppose a random sample of 40 bottles of \"CoughGone\" cough syrup was collected

Question

Suppose a random sample of 40 bottles of "CoughGone" cough syrup was collected and the alcohol content (in wt percent) of each bottle measured. A 90 % confidence interval for bottle true mean alcohol content, , was then determined from these data to be (6.6, 9.4). It was assumed that the variance of the alcohol distribution was known a.) What is the value of the sample mean, X ?(Give answer to two places past decimal.) 8 ! 160-6228 b.) What is the value of the population standard deviation, ? (Give decimal answer to two places past decimal.) 1.697 . Tries 1/5 c.) Compared to the 90 % confidence interval above, a 85% confidence interval computed from the same data would be: a wider b-narrower c same width d can't tell Tries 0/2 d.) which of the following statements best describes the meaning of a 90 % confidence interval for ? a-There is a 90 % chance that the value of is between 6.6 and 9.4 b-we can be highly confident that 90 % of all bottles of "CoughGone" cough syrup have an alcohol content between 6.6 and 9.4 wt. percent : c-If the process of selecting a random sample of size 40 and computing a 90 90 confidence interval is repeated 100 times, we expect that 90 of the resulting intervals will include ! 160-2800 e.) If the random sample size is increased from 40 to 70, what is the new width of the 90 % confidence interval? Assume the value of the population standard deviation, , remains unchanged. (Give decimal answer to two places past decimal.) 0.68 . Tries 1/5

Explanation / Answer

b) margin of error E=(9.4-6.6)/2 =1.4

for 90% CI ; z =1.6449

threrefore populations std deviation =E*n/z =1.4*40/1.6449 =5.38

c) narrower

e) new width =(9.4-6.6)*(40/70)1/2 =2.12

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