Question

Two types of valves are being tested to determine if there is a difference in pressure tolerances. Fifteen out of a random sample of $100$of Valve $A$cracked under $4500\text{psi.}$Six out of a random sample of $100$of Valve $B$cracked under $4500\text{psi}.$Test at a$5\%$ level of significance.

Short answer

Sufficient evidence is obtained to conclude that the proportion of Valve A that cracked is not equal to the proportion of Valve B that cracked.

Step-by-step solution

Given information

Here,$15$out of a random sample of $100$of Valve A cracked under $4500psi$and $6$out of a random sample of $100$of Valve B cracked under $4500psi.$Test at a $5\%$level of significance.

Calculation

The null and alternative hypothesis obtained are :

$$\begin{gathered} H_0:p_A=p_B \\ {H}_a:p_B\ne p_B \end{gathered}$$

The significant level is, $\alpha =5\%=0.05.$

The p-value is localid="1650196586213" $0.0379,$which is lower than the significant level.

Therefore, we reject the null hypothesis.

Hence, there is sufficient evidence to conclude that the proportion of Valve A that cracked is not equal to the proportion of Valve B that cracked.

Final answer

There is sufficient evidence to conclude that the proportion of Valve A that cracked is not equal to the proportion of Valve B that cracked.