find a 99 confidence interval for the difference between the

find a 99% confidence interval for the difference between the mean daily protein intakes of female vegetarians and female omnivores, x1= 39.0392, x2 = 49 9245 . s1= 18.82122, s2 = 18.96844, n1= 51, n2 = 53

Solution

CI = x1 - x2 ± t a/2 * Sqrt ( sd1 ^2 / n1 + sd2 ^2 /n2 )
Where,
x1 = Mean of Sample 1, x2 = Mean of sample2
sd1 = SD of Sample 1, sd2 = SD of sample2
a = 1 - (Confidence Level/100)
ta/2 = t-table value
CI = Confidence Interval
Mean(x1)=39.0392
Standard deviation( sd1 )=18.82122
Sample Size(n1)=51
Mean(x2)=49.9245
Standard deviation( sd2 )=18.96844
Sample Size(n1)=53
CI = [ ( 39.0392-49.9245) ±t a/2 * Sqrt( 354.2383222884/51+359.8017160336/53)]
= [ (-10.89) ± t a/2 * Sqrt( 13.73) ]
= [ (-10.89) ± 2.678 * Sqrt( 13.73) ]
= [-20.81 , -0.96]