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 43 cables and apply weights to each of them until they break. The 43 cables have a mean breaking weight of 774.3 lb. The standard deviation of the breaking weight for the sample is 15.4 lb. Find the 90% confidence interval to estimate the mean breaking weight for this type cable.

Answer 1

Answer:

At   90% confidence interval, the estimate of the mean breaking weight is (770.45, 778.15)

Step-by-step explanation:

Given that:

sample size n =43

sample mean x = 774.3

standard deviation = 15.4

confidence interval = 90%

At C.I of  90% , the level of significance ∝ = 1 - C.I

the level of significance ∝ = 1 - 0.90

the level of significance ∝ = 0.10

The critical  value for z at this level of significance is $z_{\alpha/2} = z_{0.10/2}$

$z_{0.05}$ = 1.64

The margin of error can be computed as follows:

Margin of error = $\mathtt{z_{\alpha/2} \times \dfrac{\sigma}{\sqrt{n}}}$

Margin of error =  $\mathtt{1.64 \times \dfrac{15.4}{\sqrt{43}}}$

Margin of error = $\mathtt{1.64 \times \dfrac{15.4}{6.5574}}$

Margin of error  = $\mathtt{1.64 \times2.3485}$

Margin of error = 3.8515

The mean breaking weight  for the 90% confidence interval is = $\mathtt{\overline x \pm E < \mu }$

=  $\mathtt{\overline x - E < \mu < \overline x + E}$

= ( 774.3 - 3.8515  < μ < 774.3 + 3.8515 )

= (770.4485, 778.1515)

$\simeq$ (770.45, 778.15)