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Two samples are taken with the following sample means, sizes, and standard deviations x 1 =...

Two samples are taken with the following sample means, sizes, and standard deviations

x 1 = 36 x 2 = 23

n 1 = 61 n 2 = 68

s 1 = 2 s 2 = 5

Estimate the difference in population means using a 91% confidence level. Use a calculator, and do NOT pool the sample variances. Round answers to the nearest hundredth.
_______< μ1-μ2 <________

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Answer:

Since Population variances are unknown and unequal , we use t test for difference

For unequal variances,

the degrees of freedom is

= (4/61+25/68)^2/((4/61)^2/60+(25/68)^2/67) = 89 .84 ~ 90

The critical value of t for105 df , at 9% level of significance is 1.711

The 91% CI is

= 36-23 +- 1.711*sqrt((4/61)+(25/68))

= 11.87, 14.13

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