Question

1. a simple random sample of 35 people from a certain population has a mean age of 27. the standard deviation is known to be 9.2. can we conclude that the mean age of the population is not 30? use 0.05 level of significance.

Explanation:

Step1: State the hypotheses

$H_0:\mu = 30$ (null hypothesis), $H_1:\mu
eq30$ (alternative hypothesis)

Step2: Calculate the test - statistic

The formula for the z - test statistic in a one - sample z - test is $z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}$, where $\bar{x} = 27$, $\mu = 30$, $\sigma=9.2$, and $n = 35$.
$z=\frac{27 - 30}{\frac{9.2}{\sqrt{35}}}=\frac{- 3}{\frac{9.2}{5.916}}\approx\frac{-3}{1.555}\approx - 1.93$

Step3: Determine the critical values

For a two - tailed test with a significance level of $\alpha = 0.05$, the critical values are $z_{\alpha/2}=\pm1.96$.

Step4: Make a decision

Since $-1.96<-1.93 < 1.96$, we fail to reject the null hypothesis.

Answer:

We cannot conclude that the mean age of the population is not 30 at the 0.05 level of significance.