Question

Telecorp Inc. sells holiday greeting cards via phone solicitation. Twenty-four new salespersons were hired in January 2020. The company provides training for the first month of their employment. This year, the company decided to engage a relatively expensive training consultant on a test basis to train a random sample of 12 of their 24 new hires. The other twelve received the internal training that the company provides. Management desires more information about the value of the external trainer to decide whether it will seek to extend the agreement with the training consultant to train all the new hires. The company will conduct this test in January 2020 and again in January 2021. The company does not have a directional hypothesis ex ante (in advance) regarding the impact of the trainer since it is certainly conceivable that the external training consultant could be less effective than the in-house trainers.

Sales data from year one is shown in the EXCEL data file in the tab entitled “Telecorp Inc. Sales”.

Please complete the following:

1. Produce a table of descriptive statistics for the respective groups using the EXCEL data analysis technology.
2. Highlight on the output the means, medians, and standard deviations for the two groups.
3. Based on this initial data, what is the probability that the difference in sales between the two groups is not random and instead occurred because the external trainer had an impact on the selling ability of the sales persons that received their training from her? Please provide evidence to support your conclusion.

Internal Training Group			External Training Group
Mean	21757.75		Mean	25079.25
Standard Error	1325.041098		Standard Error	1402.143798
Median	20864.5		Median	24563
Mode	#N/A		Mode	18778
Standard Deviation	4590.077007		Standard Deviation	4857.168594
Sample Variance	21068806.93		Sample Variance	23592086.75
Kurtosis	-0.869273226		Kurtosis	-0.90374431
Skewness	0.345688143		Skewness	0.255553094
Range	14665		Range	14632
Minimum	14500		Minimum	18778
Maximum	29165		Maximum	33410
Sum	261093		Sum	300951
Count	12		Count	12
Confidence Level(95.0%)	2916.395793		Confidence Level(95.0%)	3086.097691

Answer 1

Based on the given output, we are asked to find the probability that the difference in sales between the two groups is not random and instead occurred because the external trainer had an impact on the selling ability of the salespersons that received their training from her.

For a statistical test, the result is said to be significant, if the p-value (the probability of obtaining a result at least as extreme as the one obtained, when the null hypothesis is true) is less than a fixed significance level, i.e. how likely the result is due to chance. The most common level, used to denote significance, is 0.95.  And of "0.05," as in the given problem implies that the finding has a five per cent (.05) chance of not being true, which is the converse of a 95% chance of being true.

Here, we need to test:

Vs

The appropriate statistical test to test the above hypothesis would be a two-sample t-test (since the population standard deviations are unknown), with test statistic given by,

  

= -1.7217

To obtain the p-value of the test for t = -1.7217, we may use the excel function:

We get P-value = 0.0992...............(Subject to rounding error)

The p-value tells us how likely something is to be not true. Here, our test shows the probability of .0992, it means that there is a 90.08% (1-0.0992=0.9008) chance that the means of the two groups really are different.

Hence, the probability that the difference in sales between the two groups is not random and instead occurred because the external trainer had an impact on the selling ability of the salespersons that received their training from her is 90.08%...............(Subject to rounding error)

However, this probability is less than what we had fixed for, i.e 95% level. Since p-value = 0.0992 > 0.05, we fail to reject H₀ at 5% significance level; we may say that the data does not provide sufficient evidence to support the claim that the difference observed is significant.