Question

Find the mean price of gasoline in the State of Ohio. You might try this GasBuddy website. Set up a null and alternative hypothesis to see if your sample for Dayton is enough to prove that the population mean gasoline price in Dayton is different than the mean price in Ohio. Test the hypotheses. Show your work. Using a significance level of 0.05, what is your conclusion?

Dayton Mean=2.53 Standard Dev. 0.139 Sample size 15 Confidence 1.96 ME 0.034

5 CITIES IN OHIO mean 2.31 Standard Deviation 0.139

Answer 1

Assuming equal variances Since s1 / s2 = 1

Given:

For Dayton: $\bar{x_1}$ = 2.53, s₁ = 0.139, n₁ = 15

For OHIO: $\bar{x_2}$ = 2.31, s₂ = 0.139, n₂ = 5

The degrees of freedom used is n1 + n2 - 2 = 15 + 5 - 2 = 18 (since pooled variance is used)

-2 (nl - 1) * $1? + (n2 - 1) * s22 nl+n2 - 2 14 * 0.01392 + 4 * 0.01392 = 0.0193 15+5 - 2

The Hypothesis:

H0: $\mu_1$ = $\mu_2$

Ha: $\mu_1$ $\neq$$\mu_2$

This is a Two tailed test.

The Test Statistic: We use the students t test as population standard deviations are unknown.

2.53 - 2.31 T1 – T2 $**(1 + i) 70.0193+ ( + 1) = 3.07

The p Value:    The p value (2 Tail) for t = 3.07, df = 18, is; p value = 0.0066

The Critical Value:   The critical value (2 tail) at $\alpha$ = 0.05, df = , tcritical = +2.101 and -2.101

The Decision Rule:    If tobserved is > tcritical or if tobserved is < -tcritical, Then Reject H0.

Also If the P value is < $\alpha$ , Then Reject H0

The Decision:    Since t observed (3.07) is > tcritical (2.101), We Reject H0.

Also since P value (0.0066) is < $\alpha$ (0.05), We Reject H0.

The Conclusion: There is sufficient evidence at the 95% significance level to conclude that there is a difference in the mean gasoline prices in Dayton and OHIO.