Question

2. To compare prices of two grocery stores in Toronto, a random sample of items that are sold in both stores were selected and their price noted in the first weekend of July 2018: (12 Points)

Item	Store A	Store B	Difference (Store A - Store B)
1	1.65	1.98	-0.33
2	8.70	8.49	0.21
3	0.75	0.89	-0.14
4	1.05	0.99	0.06
5	11.30	11.99	-0.69
6	7.70	7.99	-0.29
7	6.55	6.99	-0.44
8	3.70	3.59	0.11
9	8.60	8.99	-0.39
10	3.90	4.29	-0.39

1. What are the null and alternative hypothesis if we want to confirm that on average, prices at Store 1 is different from the prices at Store 2, that is, the difference is different from 0?
2. What are the sample mean difference in prices and the sample standard deviation?
3. Compute the test statistic t used to test the hypothesis.
4. Compute the degree of freedom for the test statistic t.
5. Can we conclude that on average, prices at Store 1 is different from the prices at Store 2? Use the critical-value approach and α = 0.05 to conduct the hypothesis test.
6. Use the above data to construct a 95% confidence interval for the difference in prices between the two stores.

Answer 1

t-Test: Paired Two Sample for Means in excel

I performed the analysis using basic commands in MS Excel and I am giving the steps to be followed-

1. Select the Data tab and choose the Data Analysis in the top right-hand corner

2. In the Data Analysis menu choose t-Test: Paired Two Sample for Means and click OK

3. In the ‘Input Range’ box, select all the data in the columns you created, including the variable names

4. Check the ‘Labels in First Row’ box

5. In the ‘Output Range’ box, enter a cell range where Excel will place the output and click OK

6. If the p-value were less than 0.05, you would reject the null hypothesis that says the means of all categories are equal. If the p-value were greater than 0.05, then you would fail to reject the null.

Based on the above result, the question can be answered.

a)

Null hypothesis: The difference between prices at store A and store B is equal to zero

Alternate hypothesis: The difference between prices at store A and store B is not zero.

b) mean difference = -0.229

st. deviation of differences = 0.283

c) t-statistic = -2.557

d) the degree of freedom = n-1 = 10-1 = 9

where n = sample size

e) The critical value for a two-tailed test at 0.05 level of significance = 2.262

Since the t-statistic if greater than the critical value so we have enough evidence to reject the null hypothesis. We conclude that there is a significant difference between the price of store A and store B at 0.05 level of significance.

f) We have the mean, st. deviation and critical value (5%) to calculate the confidence interval

st. error of mean = se = st. deviation/ sqrt(n) =  0.283/sqrt(10) = 0.0895

confidence interval = mean +- (se * critical value )

=  0.229 +- ( 0.0895 *  2.262)

= 0.229 +- 0.2024

= (0.0266‬, 0.4314)

f)