Question

Complete the following four hypotheses, using α = 0.05 for each. The week 5 spreadsheet can be used in these analyses.

• 1. Mean sales per week exceed 42.5 per salesperson
• 2. Proportion receiving online training is less than 55%
• 3 Mean calls made among those with no training is at least 145
• 4. Mean time per call is 14.7 minutes

• 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. You can use either the p-value or the critical values to draw conclusions. Be sure to explain your conclusion.
• Compute 99% confidence intervals for the variables used in each hypothesis test, and interpret these intervals.

Answer 1

Answer:-

Given that:-

The hypothesis being tested is:

41.500	hypothesized value
44.860	mean Sales (Y)
4.733	std.dev
0.473	std.error
100	n
99	df
7.099	t
9.66E-11	p-value(one - tailed,upper)

Since the p- value (0.0000) is less than the significance level (0.05) ,we can reject the null hypothesis .

Therefore we can conclude that Mean sales per week exceed 41.5 per salesperson.

The hypothesis being tested is:

Observed	Hypothesized
0.43	0.55	p(as decimal)
43/100	55/100	p(as fraction)
43.	55.	x
100	100	n
	0.0497	std.error
	-2.41	z
	.0079	p-value(one-tailed, lower)

Since the p-value (0.0079) is less than the significance level (0.05) , we can reject the null hypothesis

Therefore ,we can conclude that the Proportion of receiving online training is less than 55%

The hypothesis being tested is

145.000	hypothesized value
152.286	mean Calls(X1)
19.038	std.dev
3.598	std .error
28	n
27	df
2.025	t
.9736	p-value (one -tailed,lower)

Since the p-value (0.9736) is greater than the significance level (0.05) ,we cannot reject the null hypothesis .

Therefore, we cannot that Mean calls made among those with no training is less than 145

The hypothesis being tested is:

15.0000	hypothesized value
15.3880	mean Time (X2)
2.8215	std dev.
0.2822	std.error
100	n
99	df
1.375	t
.0861	p-value(one -tailed, upper)

Since the p-value (0.0861) is greater than the significance level (0.05), we cannot reject the null hypothesis .

Therefore , we cannot conclude that Mean time per call is greater than 15 minutes.