Question

Consider the following hypothesis test. (a) What is your conclusion if n1-21, s12-82, n,-26, and s2 4.0? Use α 0.05 and the ρ-value approach Find the value of the test statistic. Find the p-value. (Round your answer to four decimal places.) p-value- State your conclusion. Reject Ho, we cannot conclude that σ12 # σ22 Do not reject Ho, we cannot conclude that σ1z # σ2 O Reject H , we can conclude that σ. # σ 0 D Do not reject Ho, we can conclude that σ1z # σ2 (b) Repeat the test using the critical value approach. Find the value of the test statistic. State the critical values for the rejection rule. (Round your answers to two decimal places. If you are only using one tail, enter NONE for the unused tail.) test statistic s test statistic 2 State your conclusion. Reject Ho, we cannot conclude that σ12 # σ22 Do not reject Ho, we cannot conclude that σ1z # σ2 O Reject H , we can conclude that σ. # σ 0 D Do not reject Ho, we can conclude that σ1z # σ2

Answer 1

Let $\sigma_1^2,\sigma^2_2$ be the true variances of the 2 populations.

We want to test the hypotheses

$\begin{align*} &H_0: \sigma_1^2=\sigma^2_2\leftarrow\text{null hypothesis}\\ &H_a: \sigma_1^2\ne \sigma^2_2\leftarrow\text{alternative hypothesis}\\ &\alpha=0.05\leftarrow\text{level of significance to test the hypotheses} \end{align*}$

The sample information is

$\begin{align*} &n_1=21\leftarrow\text{sample size of sample 1}\\ &s_1^2=8.2\leftarrow\text{sample variance of sample 1}\\ &n_2=26\leftarrow\text{sample size of sample 2}\\ &s_2^2=4.0\leftarrow\text{sample variance of sample 2}\\ \end{align*}$

a) the value of test statistics is

$\begin{align*} F=\frac{s_1^2}{s_2^2}=\frac{8.2}{4.0}=2.05 \end{align*}$

ans: the value of the test statistics: 2.05

The numerator degrees of freedom for this F statistics is $\begin{align*} n_1-1=21-1=20 \end{align*}$

the denominator degrees of freedom is $\begin{align*} n_2-1=26-1=25 \end{align*}$

this is a 2 tailed test (the alternative hypothesis has "not equal to"). The p-value is 2*P(F>2.05)

We can use excel function =F.DIST.RT(2.05,20,25) to get the exact value of P(F>2.05) = 0.0452

and hence the p-value=2*0.0452=0.0904

ans: The p-value =0.0904

(If we use the F tables, the closest we can get to 2.05, using the table for alpha=0.05, for df=(20,25) is 2.01. Hence the approximate p-value using the tables is 0.05*2=0.1)

We will reject the null hypothesis if the p-value is less than the level of significance alpha=0.05.

Here, the p-value of 0.0904 is greater than the significance level of 0.05. Hence we do not reject the null hypothesis.

ans: state your conclusion

$\begin{align*} \text{do not reject }H_0. \text{ We cannot conclude that }\sigma_1^2\ne\sigma^2_2 \end{align*}$

b) The value of test statistics is already calculated in part a)

ans: the value of the test statistics: 2.05

this is a 2 tailed test.

The right critical value of F for alpha=0.05 is

$\begin{align*} P(F>f_{\alpha/2})=\alpha/2=0.05/2=0.025 \end{align*}$ using  numerator degrees of freedom $\begin{align*} n_1-1=21-1=20 \end{align*}$

and the denominator degrees of freedom $\begin{align*} n_2-1=26-1=25 \end{align*}$

Using F tables for alpha=0.025, we can get f=2.30

The left critical value is obtained for $\begin{align*} P(F>f_{\alpha/2})=1-\alpha/2=1-0.025=0.975 \end{align*}$

Since we do not have an F table for 0.975, we will use the formula

$\begin{align*} F(df_n=20,df_d=25,0.975)=\frac{1}{ F(df_n=25,df_d=20,0.025)} \end{align*}$

where $\begin{align*} df_n\end{align*}$ is the numerator degrees of freedom and $\begin{align*} df_d\end{align*}$ is the denominator degrees of freedom.

Using the F table for 0.025 and closest available numerator df=24 and denominator df=20 we get 2.41

Left tail critical value of F is 1/2.41=0.4149

(Using excel function =F.INV.RT(0.025,20,25) we can get the right tail critical value of 2.30. For the left tail we can use =F.INV.RT(0.975,20,25) and get 0.4174)

We will reject the null hypothesis if the test statistics is greater than 2.30 or less than 0.4149

ans: the critical values for the rejection rule is

$\begin{align*} &\text{test statistics}\le 0.4149 \text{ (or use more accurate }0.4174)\\ &\text{test statistics}\ge 2.30\\ \end{align*}$

Here we can say that the test statistics of 2.05 is not in the rejection region.

Hence we do not reject the null hypothesis

ans: state your conclusion

$\begin{align*} \text{do not reject }H_0. \text{ We cannot conclude that }\sigma_1^2\ne\sigma^2_2 \end{align*}$