1. You need to estimate the mean number of travel days per year for salespeople.

Question

1. You need to estimate the mean number of travel days per year for salespeople. The mean of a small pilot study was 150 days, with a standard deviation of 34 days. If you must estimate the population mean within 7 days, how many salespeople should you sample? Use the 98% confidence level. (Use z Distribution Table.) (Round your answer to the next whole number.)

2. Dylan Jones kept careful records of the fuel efficiency of his new car. After the first eleven times he filled up the tank, he found the mean was 22.5 miles per gallon (mpg) with a sample standard deviation of 1.1 mpg. (Use z Distribution Table.) Compute the 95% confidence interval for his mpg. (Round your answers to 3 decimal places.)

3. The owner of Britten’s Egg Farm wants to estimate the mean number of eggs produced per chicken. A sample of 17 chickens shows they produced an average of 19 eggs per month with a standard deviation of 7 eggs per month. (Use t Distribution Table.) Develop the 98% confidence interval for the population mean. (Round your answers to 2 decimal places.)

Explanation / Answer

1) here margin of error E =7

for 98% confidence interval ; z =2.326

hence required sample size n=(Z*std deviation/E)2 =~ 128 (rounding it up)

2)

here std errror of mean =std deviaiton/(n)1/2 =1.1/(11)1/2 =0.3317

for 95% CI ; z=1.96

therefore 95% confidence interval for his mpg =sample mean -/+ z*std error =21.850 ; 23.150

3)

here std errror of mean =std deviaiton/(n)1/2 =7/(17)1/2 =1.6977

for 98% CI and (n-1=16) degree of freedom t =2.5835

therefore 95% confidence interval for his mpg =sample mean -/+ z*std error =14.61 ; 23.39

please revert for any explanation.

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