Question

Triglycerides are a form of fat found in the body. The body converts any calories it doesn’t immediately need into triglycerides which are then stored in fat cells. A researcher wants to compare triglyceride levels (measured in milligrams per deciliter) in men and women. A random sample of 44 women has

x ̄W = 104.4 mg/dL sW = 40.2 mg/dL and a random sample of 48 men has

x ̄M = 139.5 mg/dL sM = 85.3 mg/dL

(a) Find the 95% confidence interval for μW − μM the difference in mean triglyceride levels

between women and men.

(b) Is there sufficient evidence that men have a higher mean triglyceride level than women? Test at the α = 0.01 significance level.

Answer 1

a) The standard error here is computed as:

a) For n1 + n2 - 2 = 44 + 48 - 2 = 90 degrees of freedom, we get from the t distribution tables:
P( t90 < 1.987) = 0.975

Therefore the confidence interval here is obtained as:

This is the required 95% confidence interval here.

b) The test statistic here is computed as:

For 90 degrees of freedom, the p-value here is computed as:

p = P( t90 < -2.5578)

Getting it from the t distribution tables, we get:

p = P( t90 < -2.5578) = 0.0061

As the p-value here is 0.0061 < 0.01 which is the level of significance, therefore the test is significant and we can reject the null hypothesis here and conclude that men have a higher mean triglyceride level than women.