a sales manager wants to know whether a special promotional campaign is a success. he had the following data-Retail outlets :1 2 3 4 5 6 ; Sales before campaign : 50 48 31 42 28 53 ; Sales after campaign : 56 55 30 45 29 58. Test at 5% LOS.
You can probably do a two-sample t-test on your data.
Using a stat calculator, I have the following:
t= -2.78
degrees of freedom = 5
Group A: Number of items= 6
28.0 31.0 42.0 48.0 50.0 53.0
Group B: Number of items= 6
29.0 30.0 45.0 55.0 56.0 58.0
Testing at .05 level of significance using a t-table, you will likely reject the null and conclude a difference in sales.
Check this result with your own calculations.
I hope this will help.
To determine if the special promotional campaign is a success, we can perform a hypothesis test using the given data.
The null hypothesis (H0) is that the campaign had no effect on sales, and the alternative hypothesis (Ha) is that the campaign did have an effect on sales.
1. Set up the hypothesis:
- H0: The campaign had no effect on sales (μd = 0)
- Ha: The campaign had an effect on sales (μd ≠ 0)
2. Select a significance level (α):
In this case, the significance level is given as 5% or 0.05.
3. Calculate the differences in sales before and after the campaign:
Retail outlets: 1 2 3 4 5 6
Sales before campaign: 50 48 31 42 28 53
Sales after campaign: 56 55 30 45 29 58
Calculate the differences (sales after - sales before):
56 - 50 = 6
55 - 48 = 7
30 - 31 = -1
45 - 42 = 3
29 - 28 = 1
58 - 53 = 5
Sum of the differences = 6 + 7 + (-1) + 3 + 1 + 5 = 21
4. Calculate the mean of the differences:
Mean (x̄) = Sum of differences / Number of cases
Mean (x̄) = 21 / 6 = 3.5
5. Calculate the standard deviation of the differences:
Standard deviation (s) = sqrt([(Σ(x - x̄)^2) / (n - 1)])
Standard deviation (s) = sqrt([(6 - 3.5)^2 + (7 - 3.5)^2 + (-1 - 3.5)^2 + (3 - 3.5)^2 + (1 - 3.5)^2 + (5 - 3.5)^2] / (6 - 1))
Standard deviation (s) = sqrt([7.25 + 7.25 + 21.25 + 0.25 + 6.25 + 2.25] / 5)
Standard deviation (s) = sqrt(44.5 / 5)
Standard deviation (s) = sqrt(8.9)
Standard deviation (s) ≈ 2.98
6. Calculate the t-statistic:
t = (x̄ - μd) / (s / sqrt(n))
t = (3.5 - 0) / (2.98 / sqrt(6))
t ≈ 1.60
7. Determine the critical t-value:
The critical t-value is obtained from a t-table or statistical software. Since the significance level is 5% and the test is two-tailed, the critical t-value is approximately ± 2.571.
8. Compare the t-statistic with the critical t-value:
|t| < |critical t-value| implies fail to reject H0.
|1.60| < |2.571|, which means we fail to reject the null hypothesis.
9. Interpretation:
Based on the statistical test, there is insufficient evidence to conclude that the special promotional campaign had a significant effect on sales.
To determine if the special promotional campaign is statistically significant, you can use a hypothesis test called the paired t-test. The paired t-test allows you to compare the means of two related samples.
Step 1: Define the null and alternative hypothesis
The null hypothesis (H0) assumes that there is no significant difference in sales before and after the campaign. The alternative hypothesis (Ha) assumes that there is a significant difference in sales before and after the campaign.
H0: μd = 0 (There is no difference in sales before and after the campaign)
Ha: μd ≠ 0 (There is a difference in sales before and after the campaign)
Step 2: Calculate the difference in sales before and after the campaign for each retail outlet
To perform the paired t-test, you need to calculate the difference in sales for each retail outlet by subtracting the sales before the campaign from the sales after the campaign.
Outlet 1: 56 - 50 = 6
Outlet 2: 55 - 48 = 7
Outlet 3: 30 - 31 = -1
Outlet 4: 45 - 42 = 3
Outlet 5: 29 - 28 = 1
Outlet 6: 58 - 53 = 5
Step 3: Calculate the mean and standard deviation of the differences
Next, calculate the mean (x̄d) and standard deviation (sd) of the differences in sales.
x̄d = (6 + 7 - 1 + 3 + 1 + 5) / 6 = 4
sd = sqrt[((6 - 4)^2 + (7 - 4)^2 + (-1 - 4)^2 + (3 - 4)^2 + (1 - 4)^2 + (5 - 4)^2) / (6 - 1)] = 3.08
Step 4: Calculate the t-value
The t-value is calculated using the formula:
t = (x̄d - μd) / (sd / sqrt(n))
where x̄d is the mean of the differences, μd is the population mean difference (assumed to be 0 under the null hypothesis), sd is the standard deviation of the differences, and n is the sample size (which is 6 in this case).
t = (4 - 0) / (3.08 / sqrt(6)) = 2.39
Step 5: Determine the critical t-value and p-value
To determine if the t-value is statistically significant, we need to compare it to the critical t-value and calculate the p-value. The critical t-value is obtained from the t-distribution table or a statistical software based on your chosen level of significance (LOS) and degrees of freedom (df). In this case, the level of significance is 5% (0.05) and the degrees of freedom is n - 1 (6 - 1 = 5).
Using a t-distribution table or software, the critical t-value for a two-tailed test at 5% LOS with 5 degrees of freedom is approximately ±2.571.
Step 6: Compare the t-value and critical t-value, and calculate the p-value
If the absolute value of the t-value is greater than the critical t-value, there is sufficient evidence to reject the null hypothesis. Additionally, you can calculate the p-value associated with the t-value to determine its significance.
In this case, |2.39| < 2.571, which means the t-value is not greater than the critical t-value. However, you can calculate the p-value to further assess the significance.
The p-value is the probability of obtaining a t-value as extreme or more extreme than the observed t-value, assuming the null hypothesis is true. To calculate the p-value, you can use a t-distribution table or statistical software. In this case, the p-value is approximately 0.058.
Step 7: Make a conclusion
If the p-value is less than the chosen level of significance (0.05), we reject the null hypothesis and conclude that the special promotional campaign has had a significant effect on sales. However, if the p-value is greater than 0.05, we fail to reject the null hypothesis and conclude that there is no significant difference in sales before and after the campaign.
In this case, the p-value (0.058) is slightly greater than 0.05. Therefore, we fail to reject the null hypothesis. We do not have enough evidence to conclude that the special promotional campaign had a significant effect on sales.
Note: The decision to reject or fail to reject the null hypothesis is based on the chosen level of significance (LOS) and the evidence provided by the data.