Question

For a new study conducted by a fitness magazine,285 females were randomly selected. For each, the mean daily calorie consumption was calculated for a September-February period. A second sample of 275 females was chosen independently of the first. For each of them, the mean daily calorie consumption was calculated for a March-August period. During the September-February period, participants consumed a mean of 2383.5 calories daily with a standard deviation of 208. During the March-August period, participants consumed a mean of 2416.5 calories daily with a standard deviation of 262.5 . The population standard deviations of daily calories consumed for females in the two periods can be estimated using the sample standard deviations, as the samples that were used to compute them were quite large. Construct a 95% confidence interval for µ1-µ2 , the difference between the mean daily calorie consumption µ1 of females in September-February and the mean daily calorie consumption µ2 of females in March-August. Then complete the table below.

Carry your intermediate computations to at least three decimal places. Round your answers to at least two decimal places.

What is the lower limit of the 95% confidence interval?

What is the upper limit of the 95% confidence interval?

Answer 1

solution:

From the given information

For daily calorie consumption µ1 of females in September-February

  1 =2383.5 , s1=208 , n1 = 285

For daily calorie consumption µ2 of females in March-August.

   2 =2416.5 , s2= 262.5 , n2 = 275

Here, Degrees of freedom (df)=n1+n2-2 = 558

For 95% confidence level , = 1 - CL = 1 - 0.95 = 0.05

Critical value :t(/2) = t(0.025,df) = 1.964 [using t distribution table ]

The confidence interval for difference of two means is given by

CI : ( 1- 2) ± t *

:(2383.5-2416.5) ± 1.964 *

: -33 ± 39.396

: ( -72.396 , 6.396)

Therefore , the confidence interval is -72.40 < 1 - 2 < 6.40

The lower limit of the 95% confidence interval : -72.40

THe upper limit of the 95% confidence interval : 6.40