Question

In: Statistics and Probability

Complete the following four hypotheses, using α = 0.05 for each. a. Mean sales per week...

Complete the following four hypotheses, using α = 0.05 for each.

a. Mean sales per week exceeds 41.5 per salesperson

b. Proportion receiving online training is less than 55%

c. Mean calls made among those with no training is less than 145

d. Mean time per call is greater than 15 minutes

1. Using the same data set from part A, perform the hypothesis test for each speculation in order to see if there is evidence to support the manager's belief. Use the Seven Elements of a Test of Hypothesis from Section 7.1 of your text book, as well as the p-value calculation from Section 7.3, and explain your conclusion in simple terms.

2. Compute 99% confidence intervals for each of the variables described in a.-d., and interpret these intervals.

3. Write a report about the results, distilling down the results in a way that would be understandable to someone who does not know statistics. Clear explanations and interpretations are critical.

- Summary Report (about one paragraph on each of the speculations a.- d.)

- Appendix with the calculations of the Seven Elements of a Test of Hypothesis, the p-values, and the confidence intervals. Include the Excel formulas used in the calculations.

Sales (Y) Calls (X1) Time (X2) Years (X3) Type
20 210 8.0 1 NONE
32 139 16.9 4 NONE
44 165 15.7 3 ONLINE
47 186 13.5 3 ONLINE
41 180 14.0 2 ONLINE
35 150 13.0 4 ONLINE
32 120 19.9 3 NONE
46 172 14.7 3 GROUP
42 161 13.2 1 GROUP
33 143 15.4 3 NONE
42 181 11.5 4 ONLINE
55 160 17.0 3 NONE
42 140 17.5 2 GROUP
41 198 13.2 2 ONLINE
41 149 17.3 0 ONLINE
44 168 11.0 5 ONLINE
36 121 18.0 2 NONE
30 125 11.0 5 ONLINE
38 135 18.5 1 GROUP
21 185 18.9 2 ONLINE
67 155 17.9 1 NONE
45 149 13.5 1 ONLINE
52 193 13.7 5 ONLINE
37 159 18.1 0 NONE
33 152 15.0 3 GROUP
31 170 14.3 4 GROUP
44 192 16.7 1 GROUP
44 165 12.4 3 ONLINE
39 150 15.3 3 GROUP
43 174 12.7 2 ONLINE
42 168 16.4 0 ONLINE
49 178 15.1 3 ONLINE
41 164 17.8 3 GROUP
40 191 19.0 5 ONLINE
37 132 10.0 0 NONE
36 140 15.7 1 NONE
46 171 14.9 5 ONLINE
41 170 12.3 0 ONLINE
49 153 19.0 3 GROUP
42 154 14.3 2 GROUP
37 142 13.9 3 NONE
37 130 16.9 2 NONE
21 177 17.0 0 ONLINE
39 160 14.3 4 NONE
44 134 19.4 5 GROUP
49 131 14.6 1 GROUP
35 130 19.4 4 NONE
46 183 15.4 4 ONLINE
43 169 14.0 5 GROUP
41 155 16.0 2 ONLINE
48 182 13.0 2 ONLINE
39 140 12.4 1 NONE
40 157 15.4 1 ONLINE
48 167 14.8 3 ONLINE
50 144 15.8 2 NONE
44 168 12.4 2 GROUP
43 175 13.6 5 GROUP
33 150 14.9 2 GROUP
32 155 17.9 1 GROUP
46 163 16.6 2 ONLINE
48 162 14.5 4 GROUP
56 189 15.0 3 ONLINE
44 153 15.3 2 ONLINE
34 158 14.2 3 ONLINE
43 160 10.9 4 ONLINE
33 173 17.5 1 ONLINE
49 178 18.3 2 GROUP
50 189 14.3 1 ONLINE
52 184 11.4 4 ONLINE
45 174 13.6 2 ONLINE
48 188 13.6 0 ONLINE
35 149 15.6 1 GROUP
44 159 14.6 2 GROUP
44 160 14.8 2 ONLINE
67 166 18.9 1 GROUP
51 178 16.5 1 ONLINE
41 178 13.4 2 ONLINE
40 176 12.6 1 ONLINE
45 138 15.3 2 NONE
41 159 18.8 2 ONLINE
40 145 14.7 2 NONE
47 151 16.6 2 GROUP
48 186 14.2 1 ONLINE
42 194 13.6 2 ONLINE
41 152 14.5 4 GROUP
29 145 19.0 2 NONE
48 188 11.3 2 ONLINE
33 139 19.3 3 GROUP
48 201 12.5 1 ONLINE
45 156 13.2 3 GROUP
36 131 18.5 2 NONE
43 161 17.3 3 ONLINE
42 152 14.6 1 ONLINE
49 178 16.4 2 ONLINE
50 157 15.9 3 GROUP
42 154 15.3 1 GROUP
44 156 20.0 0 ONLINE
45 170 14.2 1 ONLINE
48 170 17.4 5 ONLINE
39 144 17.7 3 NONE

Solutions

Expert Solution

(a)

Sales (Y)
n= 100
mean= 42.04
sd= 7.67

here null hypothesis H0:mu=41.5 and alternate hypothesis Ha:mu>41.5 ( one tailed test)

we use standard normal variate as the sample size n=100 is more than 30 and

z=(x--mu)/(sd/sqrt(n))=(42.04-41.5)/(7.67/sqrt(100))=0.704

the one-tailed critical z(.05)=1.645 is more than calcualted z=0.704 , so we faile to reject null hypothesis and conclude that Mean sales per week does not exceeds 41.5 per salesperson

(1-alpha)*100% confidence interval for population mean=sample mean±z(alpha/2)*sd/sqrt(n)

99% confidence interval for population mean=sample mean±z(0.01/2)*sd/sqrt(n)

n= 100
sample mean= 42.04
sd= 7.67
z-value lower limit upper limit
99% confidence interval 2.58 40.06 44.02

(b)number of online=50, propotion of online=p=50/100=0.5

here null hypothesis H0:P=0.55 and althernate hypothesis Ha:P<0.55

we use z-test and z=(p-P)/sqrt(P(1-P)/n)=(0.5-0.55)/sqrt(0.55*(1-0.55)/100))=-1.005

the one-tailed critical z(.05)=1.645 is more than calcualted z=1.005(absolute value) , so we faile to reject null hypothesis and conclude that Proportion receiving online training is not less than 55%

(1-alpha)*100% confidence interval for population proportion =sample proportion (P) ±z(alpha/2)*SE(P)

99% confidence interval=P±z(0.01/2)*SE(P)=0.0.5±2.5758*sqrt(0.5*(1-0.5)/100)=0.5±0.13=(0.37,0.63)

(c) following information has been generated

Calls (X1)
n= 21.00
mean= 144.19
sd= 18.87

here null hypothesis H0:mu=145 and alternate hypothesis Ha:mu<145 ( one tailed test)

we use t-test as the sample size n=100 is more than 30 and

t=(x--mu)/(sd/sqrt(n))=(144.19-145)/(18.87/sqrt(21))=-0.197 with n-1=21-1=20 df

the one-tailed critical t(0.05,20)=1.72 is more than calcualted t=0.197 (absolute value) , so we fail to reject null hypothesis and conclude that Mean calls made among those with no training is not less than 145

(1-alpha)*100% confidence interval for population mean=sample mean±t(alpha/2,n-1)*sd/sqrt(n)

99% confidence interval for population mean=mean±t(0.01/2, n-1)*sd/sqrt(n)

n= 21
sample mean= 144.19
sd= 18.87
t-value lower limit upper limit
99% confidence interval 2.85 132.47 155.91

(d)following information has been generated using ms-excel

Time (X2)
n= 100
mean= 15.25
sd= 2.44

here null hypothesis H0:mu=15 and alternate hypothesis Ha:mu>15 ( one tailed test)

we use standard normal variate as the sample size n=100 is more than 30 and

z=(x--mu)/(sd/sqrt(n))=(15.25-15)/(2.44/sqrt(100))=1.02

the one-tailed critical z(.05)=1.645 is more than calcualted z=1.02 , so we fail to reject null hypothesis and conclude that Mean time per call is not greater than 15 minutes

(1-alpha)*100% confidence interval for population mean=sample mean±z(alpha/2)*sd/sqrt(n)

99% confidence interval for population mean=sample mean±z(0.01/2)*sd/sqrt(n)

n= 100
sample mean= 15.25
sd= 2.44
z-value lower limit upper limit
99% confidence interval 2.58 14.62 15.88

orchestra answered 2 years ago
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