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From a random sample of 17 students in an introductory finance class that uses group-learning techniques, the examination scores were found to be normally distributed with mean 20 and sample standard deviation 3. For an independent random sample of 11 students in another introductory finance class that does not use group-learning techniques, the examination scores were found to be normally distributed with mean 38 and standard deviation 2, respectively. Estimate with 90% confidence the difference between the two population mean scores (group learning score - not group learning score). Do not assume equal population variances Enter your answer as a mathematical expression (click on the answer box and then click on the yellow arrow in the answer box to get help entering an expression) Enter your answer as point estimate plus or minus margin of error rounded to four decimal places.. Preview 0.8930 * Preview -18

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Answer #1

7l n2 7l

Degree of freedom is smaller between n1 -1 or n2 -1

Therefore, here degree of freedom is 10

1.8125...........................by using TINV(0.10,10)

=>(20-38)\pm 1.8125\sqrt{\frac{9}{17}+\frac{4}{11}}

=>-18\pm 1.7128

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