A researcher wishes to estimate the average amount of money a person spends on l

Question

A researcher wishes to estimate the average amount of money a person spends on lottery tickets in a year. A sample of 45 people who play the lottery was taken. The average amount these people spend on lottery in a year was found to be $23.40. It is also known that the standard deviation of all people who purchase lottery tickets is $2. Find the 95% confidence interval for the true mean amount of money spent on lottery tickets by all people in a year. A Local restaurant wants to estimate the average amount customers spend on dinner. They sample 12 customers and found their average bill was $24.96 with a standard deviation of $2.65. Find the 80% confidence interval of the true average amount spend on dinner at this restaurant.

Explanation / Answer

4) std error of mean =std deviation/(n)1/2 =0.2981

for 95% CI, z=1.96

hence 95% confidence interval=sample mean -/+ z*std error =22.8157 ; 23.9843

5)

std error of mean =std deviation/(n)1/2 =0.765

for 80% CI and 11 degree of freedom, t=1.3634

hence 80% confidence interval=sample mean -/+ t*std error =23.9170 ; 26.0030

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