The Erie Run Outlet mall has two NIKE stores, one is on Liberty street and the other on Barnes St. Two stores laid of differently, both owners claim their stores maximizes the amounts customers will impulse purchase. A sample of 10 customers at the Liberty st store revealed they spent the following amounts: $17.58, $19.73, $12.61, $17.79, $16.22, $15.82. $15.40, $15.86, $11.82, $15.85. A sample of 14 customers from the Barnes st store revealed they spent the following amounts( more than what they intended): $18.19, 20.22, 17.38, 17.96, 23.92, 15.87, 16.47, 15.96, 16.79, 16.74, 21.40, 20.57, 19.79, 14.83.For data analysis a t-test: two sampling assuming unequal variances was used.
At the .01 sig. level is there a difference in the mean amount purchased on an impulse at the two stores? Explain these results to a person who knows about the t test for a single sample but is unfamiliar with the t test for independent means.
Hypothesis test - Independent groups
(t-TEST, unequal variance)
NIKE -Liberty st
15.868 MEAN
2.3306 SD
10 N
NIKE Barnes st
18.2921 MEAN
2.5527 SD
14 N
20 df
-2.42414 diff (liberty - barnes st)
1.00431 s err of diff
0 hypoth diff
- 2.41 t
.5502 p value two tailed
-5.28173 C I 95% lower
0.43345 C I 95% UPPER
2.85759 M.O.E
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To analyze the difference in the mean amount purchased on an impulse at the two NIKE stores, a t-test for independent means was conducted using the data provided. The t-test for independent means compares two independent groups to determine if there is a significant difference between their means.
First, let's define the null and alternative hypotheses:
Null hypothesis (H0): There is no difference in the mean amount purchased on an impulse at the two NIKE stores.
Alternative hypothesis (Ha): There is a difference in the mean amount purchased on an impulse at the two NIKE stores.
Then, let's look at the data for each store:
NIKE - Liberty St:
- Mean amount purchased on impulse: $15.868
- Standard deviation: $2.3306
- Sample size: 10
NIKE - Barnes St:
- Mean amount purchased on impulse: $18.2921
- Standard deviation: $2.5527
- Sample size: 14
Next, the formula for calculating the t-value is:
t = (mean1 - mean2) / (sqrt((sd1^2 / n1) + (sd2^2 / n2)))
Calculating the t-value:
t = (15.868 - 18.2921) / sqrt((2.3306^2 / 10) + (2.5527^2 / 14))
t ≈ -2.41
Using the t-value, we can find the p-value, which determines the statistical significance of the result. The p-value is the probability of obtaining a result as extreme or more extreme than the observed result if the null hypothesis is true.
In this case, the p-value obtained is approximately 0.5502 (two-tailed). Since the p-value is greater than the significance level of 0.01, we fail to reject the null hypothesis. This means that there is not enough evidence to conclude that there is a significant difference in the mean amount purchased on an impulse at the two NIKE stores.
Additionally, a 95% confidence interval was calculated:
- Lower bound of the 95% confidence interval: -5.28173
- Upper bound of the 95% confidence interval: 0.43345
This means that we are 95% confident that the true difference in the mean amount purchased on an impulse between the two NIKE stores lies between approximately -$5.28 and $0.43. The margin of error is approximately 2.85759.
To summarize, based on the t-test analysis, there is not enough evidence to suggest a significant difference in the mean amount purchased on an impulse at the NIKE stores on Liberty St and Barnes St. The sample data does not provide sufficient support to reject the null hypothesis.