A nutritionist wants to determine how much time nationally people spend eating a

Question

A nutritionist wants to determine how much time nationally people spend eating and drinking. Suppose for a random sample of 911 people age 15 or older, the mean amount of time spent eating or drinking per day is 1.08 hours with a standard deviation of 0.74 Determine and interpret a 95% confidence interval for the mean amount of time Americans age 15 or older spend eating and drinking each day drinking each day The nutritionist is 96% confident that the amount of time spent eating or drinking per day for any individual is between and hours

Explanation / Answer

here mean =1.08

and std deviation =0.74

sample size n=911

therefore std error =std deviation/(n)1/2 =0.0245

for 95% CI, from normal value table z =1.96

hence confidence interval =mean +/- z*std error

=1.08 +/- 1.96*0.0245

=1.0319 ; 1.1281

hence from 1.0319 to 1.1281

there is 95% probabilty that mean amount of time spent on eating and drinking will fall in above confidence interval.,

for 96% CI, z =2.0537

hence confidence interval like above 1.0296 ; 1.1304

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