Question

20. Consider the following hypothesis test: Ho: H100 A. μ A random sample of 36 observations yields a sample mean of 125. The population standard deviation is 42. Conduct the test at a - 0.01.

Answer 1

$Ho: \mu =-100$

$H_{1}:\mu \neq 100$

Level of significance is 0.01 i.e. $\alpha =0.01$

We are given

Sample mean=$\overline{x}=-125$

number of obsevations=n=36

Population standard deviation=$\sigma =42$

In this case population standard deviation is given and sample size is larger than 30, we can use normal distribution in our test.

We know calculate the standard error of mean in this case

$\sigma_{\overline{x}} =\frac{\sigma }{\sqrt{n}}=\frac{42}{\sqrt{36}}=7$

Now, we calculate the observed value of z

$z=\frac{\overline{x}-\mu _{Ho}}{\sigma _{\overline{x}}}=\frac{-125-(-100)}{7}=-3.57$

It is a two tailed test and level of significance is 0.01. We can find the value of z from normal distribution curve area tables by looking at 0.4950 area under the curve. [(100-1)/2] i.e. 2.57

Critical value of Z at significance level of 0.01 is $\pm 2.57$

In order to accept Null Hypothesis, observed value should fall between the range defined by critical value i.e (-2.57, 2.57).

Observed value of z lies outside the acceptance region. We fail to accept Null Hypothesis.

We can say that we fail to accept null hypothesis at 0.01 significance level.