Question

A random sample of n1 = 49 measurements from a population with population standard deviation σ1 = 3 had a sample mean of x1 = 12. An independent random sample of n2 = 64 measurements from a second population with population standard deviation σ2 = 4 had a sample mean of x2 = 14. Test the claim that the population means are different. Use level of significance 0.01.What distribution does the sample test statistic follow? Explain.

Answer 1

Answer:

We reject the null hypothesis that the population means are equal and accept the alternative hypothesis that the population means are different.

Step-by-step explanation:

We have large sample sizes $n_{1} = 49$ and $n_{2} = 64$, the unbiased point estimate for $\mu_{1}-\mu_{2}$ is $\bar{x}_{1} - \bar{x}_{2}$, i.e., 12-14 = -2.

The standard error is given by $\sqrt{\frac{\sigma^{2}_{1}}{n_{1}}+\frac{\sigma^{2}_{2}}{n_{2}}}$, i.e.,

$\sqrt{\frac{(3)^{2}}{49}+\frac{(4)^{2}}{64}}$ = 0.6585.

We want to test $H_{0}: \mu_{1}-\mu_{2} = 0$ vs $H_{1}: \mu_{1}-\mu_{2} \neq 0$ (two-tailed alternative). The rejection region is given by RR = {z | z < -2.5758 or z > 2.5758} where -2.5758 and 2.5758 are the 0.5th and 99.5th quantiles of the standard normal distribution respectively. The test statistic is $Z = \frac{\bar{x}_{1} - \bar{x}_{2}-0}{\sqrt{\frac{\sigma^{2}_{1}}{n_{1}}+\frac{\sigma^{2}_{2}}{n_{2}}}}$ and the observed value is $z_{0} = \frac{-2}{0.6585} = -3.0372$. Because -3.0372  fall inside RR, we reject the null hypothesis.

The test statistic follow a standard normal distribution because we are dealing with large sample sizes.