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In a sample of 57 temperature readings taken from the freezer of a restaurant, the mean is 29.6...

Question:

In a sample of {eq}57 {/eq} temperature readings taken from the freezer of a restaurant, the mean is {eq}29.6 {/eq} degrees and the population standard deviation is {eq}2.7 {/eq} degrees. What would be the {eq}80 \% {/eq} confidence interval for the temperatures in the freezer?

Confidence Interval:

The confidence interval is an estimate of the population mean. The population mean is estimated by the sample mean and the sample mean, {eq}\bar{x} {/eq} is an unbiased estimate of the population mean, {eq}\mu {/eq}.

Answer and Explanation:

Given that,

Sample size, {eq}n = 57 {/eq}

Sample mean, {eq}\bar{x} =29.6 {/eq}

Population standard deviation, {eq}\sigma = 2.7 {/eq}


The 80% confidence interval for the population mean is defined as:

{eq}\bar{x} \pm z_{0.20/2}\times \dfrac{\sigma}{\sqrt{n}} {/eq}


Excel function for the confidence coefficient:

=NORM.INV(0.2/2,0,1)


{eq}29.6 \pm 1.282\times \dfrac{2.7}{\sqrt{57}}\\ (29.1415, \ 30.0585) {/eq}


Learn more about this topic:

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Finding Confidence Intervals with the Normal Distribution

from Statistics 101: Principles of Statistics

Chapter 9 / Lesson 3
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