Question

In a sample of 28 cups of coffee at the local coffee shop, the temperatures were normally distributed with a mean of 162.5 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?

Answer 1

Answer:

The 95% confidence interval is 157.28 to 167.72

Explanation:

Number of coffee, N = 28

Mean, µ = 162.5

Standard deviation, σ = 14.1

Confidence Level, CL = 95%

z-value for 95% confidence interval = 1.96

The confidence interval is calculated below

$\begin{gathered} CI=\mu\pm z(\frac{\sigma}{\sqrt{N}}) \\ \\ CI=162.5\pm1.96(\frac{14.1}{\sqrt{28}}) \\ \\ CI=162.5\pm5.22 \\ \\ CI=(162.5-5.22)\text{ to \lparen162.5+5.22\rparen} \\ \\ CI=157.28\text{ to 167.72} \end{gathered}$

Therefore, the 95% confidence interval is 157.28 to 167.72