Q:

# A group of engineers developed a new design for a steel cable. They need to estimate the amount of weight the cable can hold. The weight limit will be reported on cable packaging. The engineers take a random sample of 41 cables and apply weights to each of them until they break. The 41 cables have a mean breaking weight of 772.3 lb. The standard deviation of the breaking weight for the sample is 15.3 lb. Find the 95% confidence interval to estimate the mean breaking weight for this type cable. ( 11.74 , 12.86 ) Your answer should be to 2 decimal places.

Accepted Solution

A:
Answer:The 95% confidence interval for the mean breaking weight for this type cable is (767.47 lb, 777.13 lb).Step-by-step explanation:Our sample size is 41The first step to solve this problem is finding our degrees of freedom, that is, the sample size subtracted by 1. So$$df = 41-1 = 40$$Then, we need to subtract one by the confidence level $$\alpha$$ and divide by 2. So:$$\frac{1-0.95}{2} = \frac{0.05}{2} = 0.025$$Now, we need our answers from both steps above to find a value T in the t-distribution table. So, with 40 and 0.025 in the two-sided t-distribution table, we have $$T = 2.021$$Now, we find the standard deviation of the sample. This is the division of the standard deviation by the square root of the sample size. So$$s = \frac{15.3}{\sqrt{41}} = 2.39$$Now, we multiply T and s$$M = T*s =2.021*2.39 = 4.83$$ThenThe lower end of the confidence interval is the mean subtracted by M. So:$$L = 772.3 - 4.83 = 767.47$$The upper end of the confidence interval is the mean added to M. So:$$L = 772.3 + 4.83 = 777.13$$The 95% confidence interval for the mean breaking weight for this type cable is (767.47 lb, 777.13 lb).