On the basis of extensive tests, the yield point of a particular type of mild steel-reinforcing bar is known to be normally distributed with $\sigma = 100$. The composition of bars has been slightly modified, but the modification is not believed to have affected either the normality or the value of $\sigma$.

a. Assuming this to be the case, if a sample of 25 modified bars resulted in a sample average yield point of 8439 lb. compute a 90% CI for the true average yield point of the modified bar.

b. How would you modify the interval in part (a) to obtain a confidence level of 92%?

#### Solution

```
sigma<-100
n<-25
xbar<-8439
alpha<-0.10
error<-qnorm(1-alpha/2)*sigma/sqrt(n)
left<-round(xbar-error,3)
right<-round(xbar+error,3)
left
```

`[1] 8406.103`

`right`

`[1] 8471.897`

Given that the sample size $n = 25$, standard deviation $\sigma = 100$ and the confidence level $0.9$.

(a) The $90$ percent confidence interval for true average yield point of the modified bar is

` $$ \begin{aligned} &\big(\overline{X}-z_{\alpha/2}\times \frac{\sigma}{\sqrt{n}},\overline{X}+z_{\alpha/2}\times \frac{\sigma}{\sqrt{n}}\big)\\ & = \big(8439-1.645\times \frac{100}{\sqrt{25}},8439+1.645\times \frac{100}{\sqrt{25}}\big)\\ &= \big(8406.103,8471.897\big) \end{aligned} $$ `

(b) The confidence level is $0.92$.

```
alpha<-0.08
error<-qnorm(1-alpha/2)*sigma/sqrt(n)
left<-round(xbar-error,3)
right<-round(xbar+error,3)
left
```

`[1] 8403.986`

`right`

`[1] 8474.014`

The $92$ percent confidence interval for true average yield point of the modified bar is

` $$ \begin{aligned} & \big(\overline{X}-z_{\alpha/2}\times \frac{\sigma}{\sqrt{n}},\overline{X}+z_{\alpha/2}\times \frac{\sigma}{\sqrt{n}}\big)\\ & = \big(8439-1.751\times \frac{100}{\sqrt{25}},8439+1.751\times \frac{100}{\sqrt{25}}\big)\\ &= \big(8403.986,8474.014\big) \end{aligned} $$ `