Question

A sample of 41 observations is selected from one population with a population standard deviation of 3.4. The sample mean is 100.0. A sample of 45 observations is selected from a second population with a population standard deviation of 5.6. The sample mean is 98.4. Conduct the following test of hypothesis using the 0.05 significance level. H0 : μ1 = μ2 H1 : μ1 ≠ μ2

a. One or two tail?

b. State the decision rule. The decision rule is to reject H0 if z is (Outside, Inside) the interval (____,____).

c. Compute value of Test statistic.

d. Reject or Do not reject?

e. what is the p value?

Answer 1

a) If we see the alternative hypothesis we see that we are conducting a bilateral test or two tail.

b) (-1.9886, 1.9886)

c)
`t=\frac{(100 -98.4)-(0)}{\sqrt{(3.4^2)/(41)}+(5.6^2)/(45)}=1.617`

d)
`p_v =2*P(t_(84)>1.617) =0.1096`

So with the p value obtained and using the significance level given
`\alpha=0.05` we have
`p_v>\alpha` so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the mean of the group 1 is NOT significantly different than the mean for the group 2.

e)
`p_v =2*P(t_(84)>1.617) =0.1096`

We can use the following excel code to calculate it: "=2*(1-T.DIST(1.617;84;TRUE))"

Explanation:

The system of hypothesis on this case are:

Null hypothesis:
`\mu_1 = \mu_2`

Alternative hypothesis:
`\mu_1 \\eq \mu_2`

Or equivalently:

Null hypothesis:
`\mu_1 - \mu_2 = 0`

Alternative hypothesis:
`\mu_1 -\mu_2\\eq 0`

Our notation on this case :

`n_1 =41` represent the sample size for group 1

`n_2 =45` represent the sample size for group 2

`\bar X_1 =100` represent the sample mean for the group 1

`\bar X_2 =98.4` represent the sample mean for the group 2

`s_1=3.4` represent the sample standard deviation for group 1

`s_2=5.6` represent the sample standard deviation for group 2

Part a

If we see the alternative hypothesis we see that we are conducting a bilateral test or two tail.

Part b

On this case since the significance level is 0.05 and we are conducting a bilateral test we have two critical values, and we need on each tail of the distribution
`\alpha/2 = 0.025` of the area.

The distribution on this cas since we don't know the population deviation for both samples is the t distribution with
`df=n_1+n_2 -2= 41+45-2=84` degrees of freedom.

We can use the following excel codes in order to find the critical values:

"=T.INV(0.025,84)", "=T.INV(1-0.025,84)"

And we got: (-1.9886, 1.9886)

Part c

The statistic is given by this formula:

`t=\frac{(\bar X_1 -\bar X_2)-(\mu_(1)-\mu_2)}{\sqrt{(s^2_1)/(n_1)}+(S^2_2)/(n_2)}`

And now we can calculate the statistic:

`t=\frac{(100 -98.4)-(0)}{\sqrt{(3.4^2)/(41)}+(5.6^2)/(45)}=1.617`

The degrees of freedom are given by:

`df=41+45-2=84`

Part d

And now we can calculate the p value using the altenative hypothesis:

`p_v =2*P(t_(84)>1.617) =0.1096`

So with the p value obtained and using the significance level given
`\alpha=0.05` we have
`p_v>\alpha` so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the mean of the group 1 is NOT significantly different than the mean for the group 2.

Part e

`p_v =2*P(t_(84)>1.617) =0.1096`

We can use the following excel code to calculate it: "=2*(1-T.DIST(1.617;84;TRUE))"