Question

In a sample of 65 temperature readings taken from the freezer of a restaurant, the mean is 31.9 degrees and the population standard deviation is 2.7 degrees. What would be the 80% confidence interval for the temperatures in the freezer?

Answer 1

Hello there. To solve this question, we have to remember some properties about confidence intervals.

To determine the confidence interval for a sample, given its mean and standard deviation, we use the formula:

`\overline{x}\pm z_(\alpha/2)\cdot(\sigma)/(√(n))`

Whereas

`\begin{gathered} \overline{x}\text{ is the mean} \\ z_(\alpha/2)\text{ is the z-score for the confidence interval} \\ \sigma\text{ is the standard deviation} \\ n\text{ is the sample size} \\ \alpha\text{ is the significance level} \end{gathered}`

The question gave:

`\begin{gathered} n=65 \\ \\ \overline{x}=31.9 \\ \\ \sigma=2.7 \\ \\ \end{gathered}`

And we want a 80% confidence interval.

We determine the significance level using the formula:

`\alpha=1-(CI)/(100)`

That is, the difference between the area from the confidence interval and the total area of the gaussian distribution:

Okay. In this case, we have that

`\alpha=1-(80)/(100)=1-0.8=0.2`

Therefore

`(\alpha)/(2)=0.1`

And we get that

`z_(\alpha/2)=z_(0.1)`

To determine this value, consider the table with Z-scores for a normal distribution that you can easily find on the internet.

In this case, we get

`z_(0.1)\approx1.28`

Such that the interval of confidence will be:

`31.9\pm1.28\cdot(2.7)/(√(65))`

Using a calculator, we find that

`31.9\pm0.43`

And the interval is:

`[31.47,\,32.33\rbrack`