Question

I would just like to know if ANOVA would be the correct test to use for the following problem below?

Fancy Fish, a fine dining upscale restaurant in Northridge, California and 2016 Open Table Diners’ Choice award winner, is enjoying its eighteenth season of providing delectable food, exceptional service, and beautiful outdoor dining experiences. “Saturday - Half-off Bottled Wine Night” has made Fancy Fish one of the San Fernando Valley’s favorite restaurants. Every Saturday night, guests can enjoy half-off every bottle of wine on the wine list while dining in the restaurant or on the terrace. The owner began offering “Saturday - Half-off Bottled Wine Night” in 2010 as an incentive for guests to dine at Fancy Fish when the economy was in a recession. Now that the economy is booming, the owner is considering whether the promotion should be continued, or even expanded. One concern is the effect that the promotion is having on the overall revenue generated from sales to the participants.

A random sample of 28 checks was collected over the course of one month of Saturday nights. Fourteen checks were from customers participating in the half-off promotion, and the other 14 checks were from customers not participating. The total revenue from each check (less alcohol, tax, and tip) is presented below. Do these data present sufficient evidence that the checks of participants is significantly different from checks of non-participants? What is your recommendation to the owner regarding the status of the promotion?

With Wine Discount	W/O Wine Discount
35	46
35	44
36	29
36	29
48	29
29	60
36	64
43	47
24	47
13	49
36	53
50	51
22	44
32	36

Answer 1

The analysis of variance (ANOVA) is used to determine whether there are any statistically significant differences between the means of three or more independent (unrelated) groups. Since here we are dealing with two groups we should apply here the independent two-sample t-test .

The sample size is n = 14.

	With Wine Discount	With Wine Discount2
	35	1225
	35	1225
	36	1296
	36	1296
	48	2304
	29	841
	36	1296
	43	1849
	24	576
	13	169
	36	1296
	50	2500
	22	484
	32	1024
Sum =	475	17381

The sample mean Xˉ is computed as follows:

Also, the sample variance s^2 is

Therefore, the sample standard deviation s is

The sample size is n = 14.

	Without Wine Discount	Without Wine Discount2
	46	2116
	44	1936
	29	841
	29	841
	29	841
	60	3600
	64	4096
	47	2209
	47	2209
	49	2401
	53	2809
	51	2601
	44	1936
	36	1296
Sum =	628	29732

The sample mean Xˉ is computed as follows:

Also, the sample variance s^2 is

Therefore, the sample standard deviation s is

The provided sample means are shown below:

Xˉ1​=33.929

Xˉ2​=44.857

Also, the provided sample standard deviations are:

s_1 = 9.864

s2​=10.96

and the sample sizes are n1​=14 and n2​=14.

(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.

(2) Rejection Region

Based on the information provided, the significance level is α=0.05, and the degrees of freedom are df = 26.

Hence, it is found that the critical value for this two-tailed test is

t_c = 2.056

The rejection region for this two-tailed test is

R={t:∣t∣>2.056}.

(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∣=2.773>tc​=2.056,

it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value is p = 0.0101,

and since p =0.0101<0.05,

it is concluded that the null hypothesis is 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 0.05 significance level.

Please like