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

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

b)
`p_v =2*P(z<-1.53)=0.126`

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

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) \\eq \mu_(2)`

The statistic would be given by:

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

Part a

Replacing we got:

`z =\frac{104-106}{\sqrt{(8.4^2)/(80) +(7.6^2)/(70)}}= -1.53`

Part b

The p value would be given by this probability:

`p_v =2*P(z<-1.53)=0.126`

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