Assume a researcher want to compare the mean alanine aminotransferase levels in two populations, individuals who...
Assume a researcher want to compare the mean alanine aminotransferase levels in two populations, individuals who drink alcohol and individuals who do not drink alcohol. The mean ALT for the individuals who do not drink alcohol is 32 with a standard deviation of 4, and 37 individuals were in the sample. The mean ALT levels for individuals who drink alcohol is 69 with a standard deviation of 19, and 38 individuals were in the sample. Construct and interpret a 95% confidence interval demonstrating the difference in means for those who drink alcohol when compared to those who do not drink. A. 4.22 and 69.78 B. 4.33 and 69.67 C. 4.32 and 69.98 D. 4.41 ad 69.59
Homework Answers
At 95% confidence interval the critical value is z0.025 = 1.96
The 95% confidence interval for 
is
(
)
+/- z0.025 * sqrt(s1^2/n1 + s2^2/n2)
= (69 - 32) +/- 1.96 * sqrt(+ (19)^2/38 + (4)^2/37)
= 37 +/- 6.18
= 30.82, 43.18
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