Question

Stainless steels are frequently used in chemical plants to handle corrosive fluids, however, these steels are especially susceptible to stress corrosion cracking in certain environments. In a sample of 295 steel alloy failures that occurred in oil refineries and petrochemical plants in Japan over the last 10 tears, 118 were caused by stress corrosion cracking and corrosion fatigue (Materials Performance, 1981). Construct a 95% confidence interval for the true proportion of alloy failures caused by stress corrosion cracking.

Answer 1

Answer: 95% confidence interval would be (0.344,0.456).

Step-by-step explanation:

Since we have given that

n = 295

x = 118

so, $\hat{p}=\dfrac{x}{n}=\dfrac{118}{295}=0.4$

At 95% confidence, z = 1.96

So, margin of error would be

$z\times \sqrt{\dfrac{p(1-p)}{n}}\\\\=1.96\times \sqrt{\dfrac{0.4\times 0.6}{295}}\\\\=0.056$

so, 95% confidence interval would be

$\hat{p}\pm \text{margin of error}\\\\=0.4\pm 0.056\\\\=(0.4-0.056,0.4+0.056)\\\\ =(0.344,0.456)$

Hence, 95% confidence interval would be (0.344,0.456).

Answer 2

Answer:

95% confidence interval would be (0.344,0.456).

Step-by-step explanation: