# Solved (Free): The true proportion of defective memory chips produced by a certain company is p. Suppose a random sample of 400 memory chips are tested

#### ByDr. Raju Chaudhari

Mar 31, 2021

The true proportion of defective memory chips produced by a certain company is p. Suppose a random sample of 400 memory chips are tested and 10 of them are found to be defective. Compute the lower limit of a 92% confidence interval for p.

#### Solution

##### Step 1 Specify the confidence level $(1-\alpha)$

Confidence level is $1-\alpha = 0.92$. Thus, the level of significance is $\alpha = 0.08$.

##### Step 2 Given information

Given that sample size $n =400$, observed value of $X$ is $X=10$.

The estimate of the proportion is $\hat{p} =\frac{X}{n} =\frac{10}{400}=0.025$.

##### Step 3 Specify the formula

$100(1-\alpha)$% confidence interval for population proportion is

 \begin{aligned} \hat{p} - E \leq p \leq \hat{p} + E. \end{aligned}

where $E=Z_{\alpha/2} \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$ and $Z_{\alpha/2}$ is the $Z$ value providing an area of $\alpha/2$ in the upper tail of the standard normal probability distribution.

##### Step 4 Determine the critical value

The critical value of $Z$ for given level of significance is $Z_{\alpha/2}$.

Thus $Z_{\alpha/2} = Z_{0.04} = 1.75$.

##### Step 5 Compute the margin of error

The margin of error for proportions is

 \begin{aligned} E & = Z_{\alpha/2} \sqrt{\frac{\hat{p}*(1-\hat{p})}{n}}\\ & = 1.75 \sqrt{\frac{0.025*(1-0.025)}{400}}\\ & =0.014. \end{aligned}

##### Step 6 Determine the confidence interval

$92$% confidence interval estimate for population proportion is

 \begin{aligned} \hat{p} - E & \leq p \leq \hat{p} + E\\ 0.025 - 0.014 & \leq p \leq 0.025 + 0.014\\ 0.0113 & \leq p \leq 0.0387. \end{aligned}

Thus, $92$% confidence interval estimate for population proportion $p$ is $(0.0113,0.0387)$.