Question

You randomly select 16 restaurants and measure the temperature of the coffee sold at each. The sample mean temperature is 162 degrees with a standard deviation of 10 degrees.
Assuming that the temperature is approximately normally distributed, a 95% confidence interval for the mean temperature is:

a) (144.7, 156.5).
b) (156.7, 167.3).
c) (168.9, 172.5).
d) (171.5, 181.3).
e) (172.6, 182.9).

Answer 1

Answer:

b) (156.7, 167.3).

Step-by-step explanation:

Sample size (n) = 16 restaurants

Mean coffee temperature (X) = 162 degrees

Standard deviation (s) = 10 degrees

degrees of freedom (n-1) = 15

t-score for a 95% confidence interval (t) = 2.131415

Assuming a normal distribution, the confidence interval is given by:

$X \pm t*\frac{s}{\sqrt{n} }\\162 \pm 1.960*\frac{10}{\sqrt{16} }$

The lower (L) and upper (U) bounds of the confidence interval are:

$L=162 - 5.3\\L=156.7\\U=162 + 5.3\\U=167.3$

The confidence interval is b) (156.7, 167.3).