The range of the values where one would expect to find the average weekly sales for the entire sales force in the company 90%.
Mean is obtained using the average formula =AVERAGE(B2:B51) which gives us an output of $3974.64. The standard deviation is also obtained using =STDEV.P(B1:B51) is 2270.121114 ≈ $2270.1211. The size of the entire sales is 50.
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The formula used is = CONFIDENCE(0.1,2270.121114,50) which gives us an interval of $ 528.06974 ≈ $528.0697. The range therefore is $3974.64 +/-$528.0697 which is $3446.5703 to $4502.7097
b. Impact of increasing the confidence level to 95%
The formula used is = CONFIDENCE(0.05,2270.121114,50) which gives us an output of $629.233907 ≈ $629.2339. The range therefore becomes $3974.64+/-$629.2339 which is $3345.4061 to $4603.8739. The impact of the increased confidence level is that it leads to an increased range as well.
c. Impact of increasing the sample size to 150 assuming the same mean and standard deviation. Confidence interval remains at 90%
The formula used is = CONFIDENCE(0.1,2270.121114,150). The value obtained is $304.881207 ≈ $304.8812 . The range becomes $3974.64 +/- $304.8812 which is $3669.7588 to $4279. 5212. The range decreases as the sample size increases.
Average weekly performance.
Rep A had a mean of $4962.266667 ≈ $4962.2667 with a standard deviation of $2029.284364 ≈ $2029.2844. The confidence interval obtained is $606.4068687 ≈ $606.4069. The range obtained by rep A is $4962.2667+/-$606.4069 giving us a range of $4355.8598 to $5568.6736. The range generated is higher than the range obtained from the population mean.
Rep B had a mean of $3909.33333 ≈ $3909.3333 with a deviation of $1622.78695 ≈ $1622.7870. The confidence interval obtained is $487.335598 ≈ $487.3356. The interval obtained is $3909.3333 +/-$487.3356 which generates a range of $3421.9977 to $4396.6689. This range is less than the range of the population mean.
a. Null and Alternative hypothesis
H 0: sale performances are statistically different
H 1 : sale performances are not statistically different
b. T-test of independent samples
From the analysis, the t Stat variable derived is > t Critical two tail. We therefore reject the null hypothesis that the means are the same and conclude that they are not the same. In comparison we observe that the two have different average weekly sales. This means that there is a statistical difference between the two reps
c. Probability value
The p-value obtained is 0.0304 which is less than the significant level of 0.1. Therefore, we reject the null hypothesis that the means of the all the reps are equal. The conclusion made is that the two are independent of each other.
4. Average weekly sales of Rep b
a. Rep B did not get promoted. The average weekly sales of the 50 reps had a mean of $3974.64 and a standard deviation of $2270.1211. Their confidence interval was $528.0697. the range obtained from it was $ 3446.5703 to $4502.7097. Rep B had a mean of $3909.3333 and standard deviation of $1622.7870. The confidence interval obtained is $487.3356, thereby giving us a range of $ 3421.9977 to $4396.6689. The sales of rep B are lesser compared to the sales of 50 reps.
b. Null and Alternative Hypotheses
H 0 : Weekly sales are greater than those of 50 reps
H 1 : Weekly sales are not greater than those of 50 reps
c. T test of independence
t Stat obtained is -1.944982655 ≈ -1.9450 which is less that the t Critical two-tail value which is 1.990847069 ≈ 1.9908. We therefore fail to reject the null hypothesis and conclude that the weekly sales of rep A are greater than the weekly sales of the other 50 reps.
d. Probability value
The p-value obtained is 0.05 and is labeled P (T<=t) two-tail which showing that the null hypothesis is true and should not be rejected.
