Question

the neighbor kids down John's block started an interesting new business selling lemonade and weapons grade plutonium. John noticed that the clientele for this burgeoning business tends to vary in the price of shoes they wear. He observed purchases of both items and recorded the current list price for the shoes using Google image searches. Assuming the prices of the shoes in question were normally distributed, use a 1% significance level to determine if there is, in fact, a difference in the mean cost of the shoes based on item purchased (use critical-values). (state the kind of test, state your null and alternate hypothesis, fill in the appropriate formulas below and clearly state the conclusion).

Lemonade	Plutonium
12.5	83.59
45.29	69.99
47.15	135.99
83.99	209.09
37.25	53.99
55.41	150
54.78	49.89
49.99	67.99
39.99	47.99
108.25	139.29
95.89	106.79
101.99	150.98
30.99	198.59
47.99	69.9
133.99	41.5
154.79
39.99
59.99
46.5
178.83
69.99
120.9

Answer 1

THESE ARE TWO INDEPENDENT GROUPS SO WE WILL USE INDEPENDENT T TEST.

TO CONDUCT THIS TEST WE WILL CHECK THE CONDITION OF EQUALITY OF POPULATION VARIANCES .

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho:μ1​ = μ2​

Ha: μ1​ ≠ μ2​

This corresponds to a two-tailed test, for which a t-test for two population means, with two independent samples, with unknown population standard deviations will be used.

Testing for Equality of Variances

A F-test is used to test for the equality of variances. The following F-ratio is obtained:

The critical values are FL​=0.288 and FU​=4.031, and since F=0.613, then the null hypothesis of equal variances is not rejected.

(2) Rejection Region

Based on the information provided, the significance level is α=0.01, and the degrees of freedom are df=35. In fact, the degrees of freedom are computed as follows, assuming that the population variances are equal:

Hence, it is found that the critical value for this two-tailed test is tc​=2.724, for α=0.01 and df=35.

The rejection region for this two-tailed test is R={t:∣t∣>2.724}.

(3) Test Statistics

Since it is assumed that the population variances are equal, the t-statistic is computed as follows:

(4) Decision about the null hypothesis: Since it is observed that ∣t∣=2.79>tc​=2.724, it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value is p=0.0085, and since p=0.00850.01, it is concluded that the null hypothesis is not rejected.

(5) Conclusion: It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population mean μ1​ is different than μ2​, at the 0.01 significance level.