# PROJECT PART B

**Topics:**Statistical hypothesis testing, Statistical inference, Statistics

**Pages:**5 (959 words)

**Published:**April 17, 2015

PROJECT PART B: Hypothesis Testing and Confidence Intervals Math 533 Applied managerial Statistics.

Instructor: Mr. Patrick Mayers.

April 12, 2015

Guillermo H. Quintela.

a. The average (mean) sales per week exceeds 41.5 per salesperson. The Null Hypothesis: The average (mean) sales per week is greater than or equal to 41.5 per salesperson. Ho:µ >= 41.5

The alternate Hypothesis: The average (mean) sales per week is less than or equal than 41.5 per salesperson. Ha:µ < 41.5

For a significance level of α = 0.05 I used the Z-Test of proportion to get the result based on the given hypothesis, being Ho:µ > 41.5. The critical p-value or this significance level is 0.022, which is less than α = 0.05. Thus, the decision rule is to reject the null Hypothesis.

Descriptive Statistics: SALES

Total

Variable Count Mean SE Mean

SALES 100 42.340 0.417

One-Sample Z

The assumed standard deviation = 4.171

95% Lower

N Mean SE Mean Bound Z P

100 42.340 0.417 41.654 2.01 0.022

N Mean SE Mean 95% CI

100 42.340 0.417 (41.522, 43.158)

Summary.

Since the value of the test statistics falls in the rejection region I decided to reject the null hypothesis. In fact, the data se provided gives me enough information to know that the null hypothesis is false, telling me that with a 95% of confidence interval or 0.05 of significance level the statement about the average (mean) sales per week is greater than 41.5 per salesperson, is true. Since the p-value for this test is smaller than the significance level of 0.05, we may concluding to reject the null hypothesis. And finally, with an 95% of confidence level in this case is giving me the right answer about the initial statement that the average (mean) sales per week exceeds 41.5 per salesperson.

b. The true population proportion of salespeople that received online training is less than 55%.

The Null Hypothesis: The average (mean) sales per week is less than 0.55. Ho:µ < 0.55

The alternate Hypothesis: The average (mean) sales per week is greater than or equal than 0.55 per salesperson. Ha:µ >= 0.55

For a significance level of α = 0.05 I used the Z-Test of proportion to get the result based on the given hypothesis, being Ho:µ < 0.55. The critical p-value or this significance level is 0.157, which is greater than α = 0.05. Thus, the decision rule is not rejecting the null Hypothesis.

Descriptive Statistics: SALES

Total

Variable TYPE Count Mean StDev

SALES GROUP 30 40.833 2.086

NONE 20 37.100 2.245

ONLINE 50 45.340 2.973

Test and CI for One Proportion

Test of p = 0.55 vs p < 0.55

95% Upper

Sample X N Sample p Bound Z-Value P-Value

1 50 100 0.500000 0.582243 -1.01 0.157

Summary.

Since the value of the test statistics is not in the critical area I will fail to reject null hypothesis. Therefore, the null hypothesis is true. I was also able to test this hypothesis using the p-value approach. Since the p-value for this test is greater than the significance level of 0.05, we may concluding do not reject the null hypothesis.

c. The average (mean) number of calls made per week by salespeople that had no training is less than 145.

Descriptive Statistics: CALLS

Total

Variable TYPE Count Mean StDev

CALLS GROUP 30 156.57 13.74

NONE 20 141.35 9.26

ONLINE 50 173.70 13.38

One-Sample T

Test of mu = 145 vs < 145

95% Upper

N Mean StDev SE Mean Bound T P

20 141.35 9.26 2.07 144.93 -1.76 0.047

The Null Hypothesis: The average (mean) number of calls made per week by salespeople that had no training...

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