Question

Consider the following hypothesis test.H0:μ1−μ2=0 Ha:μ1−μ2≠0The following results are for two independent samples taken from the two populations. Sample 1 Sample 2 n1=80n2=70 x¯¯¯1=104x¯¯¯2=106 σ1=8.4σ2=7.6a. What is the value of the test statistic?b. What is the p-value?c. With α=.05,α=.05, what is your hypothesis testing conclusion?

Answer 1

Answer:

a) $z =\frac{104-106}{\sqrt{\frac{8.4^2}{80} +\frac{7.6^2}{70}}}= -1.53$  

b) $p_v =2*P(z$

c) Since the p value is higher than the significance level provided we have enogh evidence to FAIL to reject the null hypothesis and we can't conclude that the true means are different at 5% of significance

Step-by-step explanation:

Information given

$\bar X_{1}= 104$ represent the mean for 1

$\bar X_{2}= 106$ represent the mean for 2

$\sigma_{1}= 8.4$ represent the population standard deviation for 1

$\sigma_{2}= 7.6$ represent the population standard deviation for 2

$n_{1}=80$ sample size for the group 1

$n_{2}=70$ sample size for the group 2

z would represent the statistic

Hypothesis to test

We want to check if the two means for this case are equal or not, the system of hypothesis would be:

H0:$\mu_{1}=\mu_{2}$

H1:$\mu_{1} \neq \mu_{2}$

The statistic would be given by:

$z =\frac{\bar X_1-\bar X_2}{\sqrt{\frac{\sigma^2_1^2}{n_1} +\frac{\sigma^2_2^2}{n_2}}}=$(1)

Part a

Replacing we got:

$z =\frac{104-106}{\sqrt{\frac{8.4^2}{80} +\frac{7.6^2}{70}}}= -1.53$

Part b

The p value would be given by this probability:

$p_v =2*P(z$

Part c

Since the p value is higher than the significance level provided we have enogh evidence to FAIL to reject the null hypothesis and we can't conclude that the true means are different at 5% of significance