Question

suppose you use simple random sampling to select and measure 40 backpacks weights, and find they have a mean weight of 56 ounces. assume the population standard deviation is 11.7 ounces. based on this, construct a 90% confidence interval for the true population mean backpack weight. give your answers as decimals, to two places

Explanation:

Step1: Identify the formula for confidence interval

The formula for a confidence interval for the population mean when the population standard deviation ($\sigma$) is known is: $\bar{x} \pm z_{\alpha/2} \cdot \frac{\sigma}{\sqrt{n}}$, where $\bar{x}$ is the sample mean, $z_{\alpha/2}$ is the z-score corresponding to the level of confidence, $\sigma$ is the population standard deviation, and $n$ is the sample size.

Step2: Determine the values

We have:

• Sample mean ($\bar{x}$) = 56 ounces
• Population standard deviation ($\sigma$) = 11.7 ounces
• Sample size ($n$) = 40
• Confidence level = 90%, so $\alpha = 1 - 0.90 = 0.10$, and $\alpha/2 = 0.05$. The $z$-score for $\alpha/2 = 0.05$ is $z_{0.05} = 1.645$ (from standard normal distribution tables).

Step3: Calculate the margin of error

The margin of error ($E$) is $z_{\alpha/2} \cdot \frac{\sigma}{\sqrt{n}}$.
Substitute the values:
$E = 1.645 \cdot \frac{11.7}{\sqrt{40}}$
First, calculate $\sqrt{40} \approx 6.3246$
Then, $\frac{11.7}{6.3246} \approx 1.85$
Then, $E = 1.645 \cdot 1.85 \approx 3.04$ (rounded to two decimal places)

Step4: The confidence interval is $\bar{x} \pm E$

So, the first box is the sample mean, 56.00, and the second box is the margin of error, 3.04.

Answer:

56.00 $\pm$ 3.04