Question

Two-sample T for LeanBodyMass

Sex N Mean StDev SE Mean
Male 75 52.63 6.66 0.77
Female 75 43.42 6.05 0.70

Difference = μ (Male) - μ (Female)
Estimate for difference: 9.21
95% CI for difference: (-, -)
T-Test of difference = 0 (vs ≠): T-Value = - P-Value = - DF = -
Both use Pooled StDev = 6.3633





Two-sample T for MetRate
SE
Sex N Mean StDev Mean
Male 75 1626 227 26
Female 75 1258 172 20

Difference = μ (Male) - μ (Female)
Estimate for difference: 367.9
95% CI for difference: (-, -)
T-Test of difference = 0 (vs ≠): T-Value = - P-Value = - DF = -
Both use Pooled StDev = 201.5088

b. Calculate a 95% confidence interval to estimate the difference between the average lean body mass of adolescent males and the lean body mass of adolescent females

Answer 1

Answer on Question #55538 – Math – Statistics and Probability

Two-sample T for LeanBodyMass

Difference = μ (Male) - μ (Female)

Estimate for difference: 9.21

95% CI for difference: (-, -)

T-Test of difference = 0 (vs ≠): T-Value = - P-Value = - DF = -

Both use Pooled StDev = 6.3633

b. Calculate a 95% confidence interval to estimate the difference between the average lean body mass of adolescent males and the lean body mass of adolescent females

Solution

We have $75 + 75 - 2 = 148$ degrees of freedom.

$$T^* = t_{148, \frac{1 - 0.95}{2}} = t_{148, 0.025} = 1.9761.$$

A 95% confidence interval to estimate the difference between the average lean body mass of adolescent males and the lean body mass of adolescent females is

$$CI = \left(9.21 - 1.9761 \cdot 6.3633 \sqrt{\frac{1}{75} + \frac{1}{75}}; 9.21 + 1.9761 \cdot 6.3633 \sqrt{\frac{1}{75} + \frac{1}{75}}\right) = (7.157; 11.263)$$

Answer: (7.157; 11.263).

Two-sample T for MetRate

Difference = μ (Male) - μ (Female)

Estimate for difference: 367.9

95% CI for difference: (-, -)

T-Test of difference = 0 (vs ≠): T-Value = - P-Value = - DF = -

Both use Pooled StDev = 201.5088

Solution

We have $75 + 75 - 2 = 148$ degrees of freedom.

$$T^* = t_{148, \frac{1 - 0.95}{2}} = t_{148, 0.025} = 1.9761.$$

A 95% confidence interval to estimate the difference between the average MetRate of males and females is

$$\begin{array}{l} Cl = \left(367.9 - 1.9761 \cdot 201.5088 \sqrt{\frac{1}{75} + \frac{1}{75}}; 367.9 + 1.9761 \cdot 201.5088 \sqrt{\frac{1}{75} + \frac{1}{75}}\right) \\ = (302.87; 432.93) \end{array}$$

Answer: (302.87; 432.93).

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