A software engineer is creating a new computer software program. She wants to make sure that the crash rate is extremely low so that users would give high satisfaction ratings. In a sample of $400$ users, $20$ of them had their computers crash during the $1$-week trial period.

$(a)$ What is $\hat{p}$?


$(b)$ What is the $95$% confidence interval for $\hat{p}$? (Use a table or technology. Round your answers to three decimal places.)

$$(0.0286 , 0.0714)$$

I don't understand. Can someone please explain how $(a)$ and $(b)$ were achieved?

Thank you


The best estimate for $p$, $\hat{p}$, is obtained by checking the proportion in your sample. Thus the best you can do is $\hat{p} = \frac{20}{400} = \frac{1}{20}$.

Confidence intervals for a proportion are generated by

$$ \hat{p} \pm z^* \cdot \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} $$

where $z^*$ is the critical value for the desired confidence interval. Since we desire a $95\%$ confidence interval, $ z^* = 1.96$ here (remember this magic number! This pops up time and time again. Other magic numbers are $1.645$ at the $90\%$ level and $2.575$ for the $99\%$ level). Use the value of $\hat{p}$ above and the sample size to generate the interval.

I'm almost certain that your notes have this information. Can you take it from here?


$\hat{p}$ is the proportion of success. Since we have $20$ successes in $400$ trials we get


A $(100-\alpha)$% confidence interval for the population proportion $p$ is given by

$$\hat{p}\pm z_{\alpha/2}\cdot\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$$

where $z_{0.05/2}\approx1.96$. You can find this by looking at a $z$-table and finding what $z$-score yields

$$P(Z\leq z)=0.975$$

since $1-0.975=0.025=\frac{0.05}{2}$

In R statistical software, we can get a more accurate value

> qnorm(.975)
[1] 1.959964

We then get

$$\hat{p}\pm 1.96\cdot\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$$

Plug in $\hat{p}$ and $n$ and you will obtain the desired confidence interval.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.