Question: Karen wants to advertise how many chocolate chips are in each Big Chip cookie at her bakery. She randomly selects a sample of 76 cookies and finds that the number of chocolate chips per cookie in the sample has a mean of $11.5$ and a standard deviation of $2.6$. What is the $90\%$ confidence interval for the number of chocolate chips per cookie for Big Chip cookies? Enter your answers accurate to one decimal place (because the sample statistics are reported accurate to one decimal place).
1. **State the problem:** We want to find the $90\%$ confidence interval for the population mean $\mu$ of chocolate chips per cookie based on a sample of size $n=76$, sample mean $\bar{x}=11.5$, and sample standard deviation $s=2.6$.
2. **Formula for confidence interval:** Since the sample size is large ($n>30$), we use the $z$-distribution.
$$\text{CI} = \bar{x} \pm z_{\alpha/2} \times \frac{s}{\sqrt{n}}$$
where $z_{\alpha/2}$ is the critical value for $90\%$ confidence.
3. **Find critical value:** For $90\%$ confidence, $\alpha=0.10$, so $\alpha/2=0.05$. The $z$-score for $0.05$ in each tail is approximately $1.645$.
4. **Calculate standard error:**
$$\text{SE} = \frac{s}{\sqrt{n}} = \frac{2.6}{\sqrt{76}}$$
Calculate $\sqrt{76} \approx 8.7178$.
$$\text{SE} = \frac{2.6}{8.7178} \approx 0.2981$$
5. **Calculate margin of error:**
$$\text{ME} = z_{\alpha/2} \times \text{SE} = 1.645 \times 0.2981 \approx 0.4904$$
6. **Calculate confidence interval:**
$$\text{Lower bound} = 11.5 - 0.4904 = 11.0096 \approx 11.0$$
$$\text{Upper bound} = 11.5 + 0.4904 = 11.9904 \approx 12.0$$
7. **Answer:**
$$11.0 < \mu < 12.0$$
This means we are $90\%$ confident that the true mean number of chocolate chips per cookie is between $11.0$ and $12.0$.