I have a problem on statistics 5.Data on household vehicle miles of travel (VMT) are compiled annually by the Federal Highway Administration and are published in National Household Travel Survey, Summary of Travel Trends. Independent random samples of 15 midwestern households and 14 southern households provided the following data on last year’s VMT, in thousands of miles. At the 5% significance level, does there appear to be a difference in last year’s mean VMT for midwestern and southern households? Use both p-value and critical value approach. Assume population variance to be equal

Midwest
16.2 , 14.6 , 11.2 , 24.4 , 9.4 12.9 , 18.6 , 16.6 , 20.3 ,15.1 , 17.3 , 10.8 , 16.6 , 20.9 , 18.3
South
22.2, 19.2 , 9.3 , 24.6 ,20.2 , 15.8, 18.0 , 12.2 , 20.1 , 16.0 , 17.5 , 18.2 , 22.8 , 11.5

Step-by-step explanation:

Hello!

The objective is to  compare the VMT of mid western households and southwestern households. For this two independent random samples of households from both areas and their VMT were recorded:

Be

X₁: VMT of a mid western household

Midwest

16.2 , 14.6 , 11.2 , 24.4 , 9.4 12.9 , 18.6 , 16.6 , 20.3 ,15.1 , 17.3 , 10.8 , 16.6 , 20.9 , 18.3

n₁= 15

∑X₁= 243.20

∑X₁²= 4175.98

X[bar]₁= 16.21

S₁²= 16.64

S₁= 4.08

X₂: VMT of a southwestern household

22.2, 19.2 , 9.3 , 24.6 ,20.2 , 15.8, 18.0 , 12.2 , 20.1 , 16.0 , 17.5 , 18.2 , 22.8 , 11.5

n₂= 14

∑X₂= 247.60

∑X₂²= 4633.24

X[bar]₂= 17.69

S₂²= 19.56

S₂= 4.42

The parameters of study are the population means, if the claim is that the VMT of households is different in both areas, then you'd expect the population means to be different too.

The hypotheses are:

H₀: μ₁ = μ₂

H₁: μ₁ ≠ μ₂

α: 0.05

Assuming both populations are normal and since both population variances are equal the test to apply is an independent samples t test pooled variance:

Sa= 4.249= 4.25

Critical value approach:

This test is two-tailed, this means that the rejection region is divided in two tails:

The decision rule is:

If ≤ -2.052 or if ≥ 2.052, reject the null hypothesis.

If -2.052 < < 2.052, do not reject the null hypothesis.

The calculated value is within the "no rejection region" so the decision is to not reject the null hypothesis.

Using the p-value approach:

The p-value is the probability of obtaining a value as extreme as the calculated value of the statistic under the null hypothesis (). Just as the significance level, the p-value is two tailed, you can calculate it as:

P(t₂₇ ≤ -0.93) + P(t₂₇ ≥ 0.93)= P(t₂₇ ≤ -0.93) + (1 - P(t₂₇ < 0.93)= 0.1796 + ( 1 - 0.8204)= 0.1796*2= 0.3592

p-value= 0.3592

The p-value is always compared to the significance level, the decision rule for this approach is:

If the p-value ≤ α, reject the null hypothesis.

If the p-value > α, do not reject the null hypothesis.

The p-value is greater than α, so the decision is to not reject the null hypothesis.

At a 5% significance level, there is no significant evidence to reject the null hypothesis. You can conclude that the population means of the VMT for households of the Midwest  South ers households.

I hope this mhelps!

282 6/7 cm^2

step-by-step explanation:

24/56=hx^7

step-by-step explanation: