Assume a researcher wants to compare the mean Alanine Aminotransferase (ALT) lev

Question

Assume a researcher wants to compare the mean Alanine Aminotransferase (ALT) levels in two populations, individuals who drink alcohol and individuals who do not drink alcohol. The mean ALT levels for the individuals who do not drink alcohol is 32 with a standard deviation of 14, 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 individuals who drink alcohol when compared to those who do not drink alcohol. Possible Answers: A. The researchers are 95% confident that the true mean difference in ALT values between the population of drinkers and population of non-drinkers is between 24.33 and 39.67. B. The researchers are 95% confident that the true mean difference in ALT values between the population of drinkers and population of non-drinkers is between 24.41 and 39.59. C. The researchers are 95% confident that the true mean difference in ALT values between the population of drinkers and population of non-drinkers is between 24.22 and 39.78. D. The researchers are 95% confident that the true mean difference in ALT values between the population of drinkers and population of non-drinkers is between 24.32 and 39.68.

Explanation / Answer

x1 = 69 , s1 = 19 , n1 = 38
x2 = 32 , s2 = 14 , n2 = 37

z value at 95% = 1.96

CI = (x1 - x2) + / - z * sqrt(s1^2/n1 + s2^2/n2)
= ( 69-32) +/- 1.96 * sqrt(19^2/38 + 14^2/37)
= 29.46, 44.54

Possible Answers should be. The researchers are 95% confident that the true mean difference in ALT values between the population of drinkers and population of non-drinkers is between 29.46 and 44.54

this possible answer is not given in option please have a look

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