Question

2.) The Kellogg Company periodically compares sales across departments. In one particular instance, they would like to determine if sales of snacks are different from sales of frozen foods. If snacks are group 1 and frozen foods are group 2, what are the hypotheses for this test?

	1) HO: μ1 ≤ μ2 HA: μ1 > μ2
	2) HO: μ1 ≥ μ2 HA: μ1 < μ2
	3) HO: μ1 = μ2 HA: μ1 ≠ μ2
	4) HO: μ1 ≠ μ2 HA: μ1 = μ2
	5) HO: μ1 > μ2 HA: μ1 ≤ μ2 2,) In a consumer research study, several Meijer and Walmart stores were surveyed at random and the average basket price was recorded for each. You wish to determine if the average basket price for Meijer is different from the average basket price for Walmart. It was found that the average basket price for 19 randomly chosen Meijer stores (group 1) was $64.761 with a standard deviation of $10.6264. Similarly, a random sample of 25 Walmart stores (group 2) had an average basket price of $64.728 with a standard deviation of $13.241. Perform a two independent samples t-test on the hypotheses Null Hypothesis: μ1 = μ2, Alternative Hypothesis: μ1 ≠ μ2. What is the test statistic and p-value of this test? You can assume that the standard deviations of the two populations are statistically similar. 1) Test Statistic: -0.009, P-Value: 0.9929 2) Test Statistic: 0.009, P-Value: 0.5035 3) Test Statistic: 0.009, P-Value: 0.4964 4) Test Statistic: 0.009, P-Value: 0.9929 5) Test Statistic: 0.009, P-Value: 1.5035

Answer 1

Here we have given that

Claim: To check weather the sales of snacks group-1 are different from sales of frozen foods group-2.

The hypothesis is

v/s

That is here option C is Correct

Now, we have given that

n1=1^st sample size group 1= 19

=1^st sample mean group 1 =64.761

S1=1^st sample standard deviation group1 =10.6264

n2 =2^nd sample size group 2=25

= 2^nd sample mean group 2=64.728

S2=2^nd sample standard deviation group 2=13.241

Here we assume the two population variances are equal

We can find the test statistics

Now,

We find the test statistics for two sample t test

Test statistics is:

Where =pooled sample variance

Now we find it

=106.84

Now we get test statistics is

= 0.009   0.01

here we get the

Test statistics is 0.009

Now we find the p-value

=level of significance=0.05

Degrees of freedom = n1+n2-2 = 19+25-2 =42

This is two tailed test

we get

P-value= 0.9929 using EXCEL =TDIST(TSTAT=0.009 , D.F=42,TAIL=2)

We get

P-value=0.9929

That is here option D (4) is correct

Decision:

P-value > 0.05

That is we fail to reject Ho Null hypothesis

conclusion:

That is we say that there is no sufficient evidence that the sales of snacks group-1 are different from sales of frozen foods group-2.