Question

A researcher is interested in estimating the mean weight of a semi trailer truck to determine the potential load capacity. She takes a random sample of 17 trucks and computes a sample mean of 20,000 pounds with sample standard deviation of 1,500. The 95% confidence interval for the population mean weight of a semi trailer truck is ______________.

Answer 1

Answer: 95% confidence interval = 20,000 ± 2.12$\times$$\frac{1500}{\sqrt{17} }$

( 19228.736 , 20771.263 ) OR ( 19229 , 20771 )

Step-by-step explanation:

Given :

Sample size(n) = 17

Sample mean = 20000

Sample standard deviation = 1,500

5% confidence

∴ $\frac{\alpha}{2}$ = 0.025

Degree of freedom ($d_{f}$) = n-1 = 16

∵ Critical value at ( 0.025 , 16 ) = 2.12

∴ 95% confidence interval = mean ± $Z_{c}$$\times$$\frac{\sigma}{\sqrt{n} }$

Critical value  at 95% confidence interval = 20,000 ± 2.12$\times$$\frac{1500}{\sqrt{17} }$

( 19228.736 , 20771.263 ) OR ( 19229 , 20771 )