Answer:
A 99% confidence interval for the mean breaking strength of blocks produced is
![[959.987, 1011.213]](https://img.qammunity.org/2020/formulas/mathematics/college/slmfy9126txsq31pak798spp8zfqksr3av.png)
Explanation:
A (1 -
)x100% confidence interval for the average break in these conditions It is an interval for the population mean with unknown variance and is given by:
![[\bar x -T_{(n-1,(\alpha)/(2))} (S)/(√(n)), \bar x +T_{(n-1,(\alpha)/(2))} (S)/(√(n))]](https://img.qammunity.org/2020/formulas/mathematics/college/5yo5b8pey2r7pj0v58cadnvd1xda01aplk.png)
![\bar X = 985.6psi](https://img.qammunity.org/2020/formulas/mathematics/college/mqywh2ca9aiwhiormtxt9pm0p9vrsy2qy4.png)
![n = 9](https://img.qammunity.org/2020/formulas/mathematics/middle-school/36vd2fwr1lh5ni0je9y7m2x44nvkz3idu0.png)
![\alpha = 0.01](https://img.qammunity.org/2020/formulas/mathematics/college/kqd5yr8wwm6794o8hurexgrd4rr1z1s672.png)
![T_{(n-1,(\alpha)/(2))}=3.355](https://img.qammunity.org/2020/formulas/mathematics/college/5ps69uoi1858jcman5ybrgiork45z2kh13.png)
![S = 22.9](https://img.qammunity.org/2020/formulas/mathematics/college/dsqlsz4f1wtgrcyojjifcv74obz05d50um.png)
With this information the interval is determined by:
![[985.6 - 3.355(22.9)/(√(9)), [985.6 - 3.355(22.9)/(√(9))] = [959.987, 1011.213]](https://img.qammunity.org/2020/formulas/mathematics/college/20pkt8rrdytpopm9nxaz6b66kdn8yyhjmb.png)