You measure 30 backpacks' weights, and find they have a mean weight of 73 ounces. Assume the population standard deviation is 7.1 ounces. Based on this, construct a $95\mathrm{\%}$ confidence interval for the true population mean backpack weight.

Give your answers as decimals, to two places
$$\pm$$
ounces
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Step 1 ：Given that the sample mean ($\overline{x}$) is 73 ounces, the population standard deviation ($\sigma$) is 7.1 ounces, and the sample size ($n$) is 30.

Step 2 ：The Z-score for a 95% confidence interval is approximately 1.96.

Step 3 ：Substitute these values into the formula for a confidence interval: $\overline{x}\pm Z\frac{\sigma }{\sqrt{n}}$

Step 4 ：Calculate the margin of error: $1.96\times \frac{7.1}{\sqrt{30}}\approx 2.54$ ounces

Step 5 ：Subtract the margin of error from the sample mean to get the lower bound of the confidence interval: $73-2.54=70.46$ ounces

Step 6 ：Add the margin of error to the sample mean to get the upper bound of the confidence interval: $73+2.54=75.54$ ounces

Step 7 ：Final Answer: The 95% confidence interval for the true population mean backpack weight is approximately $\boxed{{\displaystyle 70.46}}$ ounces to $\boxed{{\displaystyle 75.54}}$ ounces.