You measure 50 textbooks' weights, and find they have a mean weight of 77 ounces. Assume the population standard deviation is 12.3 ounces. Based on this, construct a 95% confidence interval for the true population mean textbook weight.Give your answers as decimals, to two places < μ<

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Given that:

- You measure 50 textbooks' weights.

- They have a Mean of 77 ounces.

- The Population Standard Deviation is 12.3 ounces.

You need to use the following formula:

[tex]CI=\bar{x}\pm z\frac{\sigma}{\sqrt{n}}[/tex]

Where:

- The Sample Mean is:

[tex]\bar{x}[/tex]

- The z-value for the corresponding Confidence Interval Level is "z".

- The Sample Standard Deviation is σ.

- The Sample Size is "n".

In this case:

[tex]\begin{gathered} \bar{x}=77 \\ \sigma=12.3 \\ n=50 \end{gathered}[/tex]

By definition, for a 95% Confidence Interval:

[tex]z=1.96[/tex]

Then, by substituting values and evaluating, you get these two values:

[tex]CI=77+1.96\cdot\frac{12.3}{\sqrt{50}}\approx80.41[/tex]

[tex]CI=77-1.96\cdot\frac{12.3}{\sqrt{50}}\approx73.59[/tex]

Hence, the answer is:

[tex]73.59<\mu<80.41[/tex]