Question

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 47 cables and apply weights to each of them until they break. The 47 cables have a mean breaking weight of 777.4 lb. The standard deviation of the breaking weight for the sample is 15.5 lb. Find the 90% confidence interval to estimate the mean breaking weight for this type cable. Round to 2 decimals

Answer 1

Answer :

We have given :

n = sample size = 47

x̄ = sample mean  = 777.4 lb

s = sample standard deviation  = 15.5 lb

α = 1 - 0.90 = 0.10

## Find the 90% confidence interval to estimate the mean breaking weight for this type cable.

Round to 2 decimals

μ = [  x̄  ± ( t critical value * standard error ) ]

# t critical value = t ( α /2 , df ) here df

= degree of freedom = n -1 = 46  

t ( α /2 , df ) = t ( 0.10 /2 , 46 )

=  ± 1.6787 ( use statistical table )

# standard error = s / √ n = 15.5 /  √ 47

= 2.2609

μ = [  x̄  ± t critical value * standard error ]

= [ 777.4 ± ( 1.6787 * 2.2609 ) ]

= [ 777.4   ± 3.7954 ]

lower limit =  777.4 -  3.7954 = 773.6046

ie 773.60

upper limit = 777.4 + 3.7954 = 781.1954

ie 781.20

Interpretation :

we can say 90 % confident that estimate the mean breaking weight for this type cable is lies

within 773.60 and 781.20