Question

The fasting blood sugar (FBS) of 15 randomly selected patients is given below:

110, 118, 130, 140, 142, 146, 112. 100, 95, 98, 96, 122, 123, 124, 130

 Do you think that the mean FBS of the patients in the population is 110? It is known that the population variance of FBS is 300.

Answer 1

$n=15 \\ \sigma^2=300 \\ \sigma = 17.32 \\ \bar{x}= \frac{110+ 118+ ...+ 124+ 130}{15} = 119.06 \\ H_0: \mu = 110 \\ H_1: \mu ≠ 110$

Test-statistic:

$Z= \frac{\bar{x} -\mu}{\sigma / \sqrt{n}} \\ Z = \frac{119.06 -110}{17.32 / \sqrt{15}} = 2.025$

Let use 1% significant level.

Two-tailed test.

Reject H₀ if Z≤ -2.575 or Z>2.575.

$Z=2.025<Z_{crit}= 2.575$

Accept H₀.

There is enough evidence to conclude that the mean FBS of the patients in the population is 110 at 1% significant level.