In a sample of 28 cups of coffee at the local coffee shop, the temperatures were...

In a sample of 28 cups of coffee at the local coffee shop, the temperatures were normally distributed with a mean of 182.5 degrees with a sample standard deviation of 14.1 degrees. What would be the 95% confidence interval for the temperature of your cup of coffee?

Solutions

Expert Solution

Solution :

Given that,

= 182.5

s = 14.1

n = 28

Degrees of freedom = df = n - 1 = 28 - 1 = 27

At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

t_/2,df = t_0.025,27 =2.052

Margin of error = E = t_/2,df * (s / $\sqrt$ n)

= 2.052 * (14.1 / $\sqrt$ 28 )

= 5.5

Margin of error = 5.5

The 95% confidence interval estimate of the population mean is,

- E < < + E

182.5 - 5.5 < < 182.5 + 5.5

177 < < 188

The temperature of cup of coffee IS ( 177 , 188 )

orchestra answered 1 year ago

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