In comparing the times until failure in hours of two differe

In comparing the times until failure (in hours) of two different types of light bulbs, we obtain the sample characteristics n1 = 45, x = 984, s2x = 8,742 and n2 = 52, y = 1,121, s2x = 9,411. Find an approximate 90% confidence interval for the difference of the two population means.

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)=984
Standard deviation( sd1 )=93.499
Sample Size(n1)=45
Mean(x2)=1121
Standard deviation( sd2 )=97.01
Sample Size(n1)=52
CI = [ ( 984-1121) ±t a/2 * Sqrt( 8742.063001/45+9410.9401/52)]
= [ (-137) ± t a/2 * Sqrt( 375.25) ]
= [ (-137) ± 1.68 * Sqrt( 375.25) ]
= [-169.54 , -104.46]