7. A research firm conducted a survey to determine the average amount college st

Question

7. A research firm conducted a survey to determine the average amount college students spend on beer & wine during a week. A random sample of 48 college students revealed that mean of $45.00 with a standard deviation of $12.78.   Construct 95% confidence interval on the population mean.

a. If we construct 90% confidence interval rather than 95%, than CI be wider or narrower from above answer. Why?

b. If we take 75 samples rather than 48 samples, than CI will be wider or narrower from above answer. Why?

Explanation / Answer

7) here std error of mean =std deviation/(n)1/2 =12.78/(48)1/2 =1.8446

for 95% CI; z=1.96

therefore 95% confidence interval =sample mean -/+ z*std error =41.3846 ; 48.6154

a)

for 90% CI; value of z =1.645

as CI length is proportional to z which is less for 90% CI; therefore CI will be narrower.

b) as CI length is proportional to std errror which is inversely proportional to sample size.

therefore as sample size increases std error will decrease and therefore CI will decreases ,

HEnce  CI will be narrower

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