Question

Problem 22) For a project that investigates the average weight difference by gender of patients from the Mayo clinic, the researcher collected a sample of 64 male patients and another sample of 36 female patients. The researcher then calculated the sample mean of male patients as 160 lbs with a sample standard deviation of 20 lbs, and the sample mean of female patients as 140 lbs with a sample standard deviation of 10 lbs. Test the claim that the average weight of male patients is more than that of female patients. Assume unequal population standard deviation. Given the significance level equals 5%. (Question). Step 2: Find the Critical Value.

Answer 1

$H_0: \mu_m=\mu_f$

$H_a: \mu_m>\mu_f$

Test statistic > T-critical value

A E F H 1 к L M 1 2 3 с D difference between mean assuming that the variance is unequal a) Mean SD Male 160 20 Female 140 10 n formula used 4 64 5 36 6 7 8 9 We shall be using T statistic since the population standard deviation is not given for both the samples Test statistict = 6.6564 ((C4-C5))/SQRT (D412/E4)+(D542/E5)) df = 96.97 ((D442/E4)+(D542/E5))^2/((D444/(E42*(E4-1)))+(D514/(E512*(E5-1)))) Significance level 0.05 T critical 1.6609 .-T.INV(C10,09) P value for right tailed test 0.0000 10 11 12

Hence we will reject the null hypothesis.

Thus we have evidence that the average weight of male patients is more than that of female patients.