Question

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 σ = 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 σ. (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) Compute a 92% CI for the true average yield using the sample data in (a).

Answer 1

Answer:

Step-by-step explanation:

Given that the yield point of a particular type of mild steel-reinforcing bar is known to be normally distributed with σ = 100

Modified mean without change of sigma = 8439

Sample size n =25

Std error of sample = $\frac{\sigma}{\sqrt{n} } \\=1.2$

For 90% confidence interval we use Z critical value since we know population std deviation

Margin of error = $1.645(1.2)\\= 1.974$

a) Confidence interval

=$(8439-1.974, 8439+1.974)\\= (8437.026, 8440.974)$

b)For 92% critical value changes to 1.75

Confidence interval

=$(8439-1.78(1.2), 8439+1.78(1.2))\\=(8436.864, 8441.136)$