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Accounting, Finance, SPSS

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# Stat 2606 E Assignment 4 (Statistics Project Sample)

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The Stat 2606 E Assignment 4 paper contained basic questions regarding hypothesis test. It involved stating the null and alternative hypothesis and computing the relevant statistics for undertaking the test. Upon computation, one would conclude wither the null hypothesis is accepted or rejected at 5 % significance level. I also compared different methods in undertaking the hypothesis tests. source..

Content:

Student’s Name
Professor’s Name
Course
Date
Stat 2606 E: Assignment 4
Question 1
n = 25
x=9.48
µ =10
σ=1.43
H0: true mean net weight is equal to the net weight of 10 ounces indicated on the container
H1: true mean net weight is lower than the net weight of 10 ounces indicated on the container
t = 9.48-101.4325
= 2.1742
Critical value = 2.064
Since the test statistic 2.1742 is greater than the critical value 2.064, we reject the null hypothesis that true mean net weight is lower than the net weight of 10 ounces indicated on the container.
The test is valid if there are no outliers and n is at least 15.
Question 2
H0: p=0.5
H1: p>0.5
n = 100
p = 58100
=0.58
z = 0.58-0.50.5*0.5100
= 0.080.05
=1.6
P-value = 0.9452007
Critical value = 1.645
(i) The p-value method
Since the p-value is greater than 0.05, we fail to reject the null hypothesis that the true proportion, p, of cola drinkers that prefer Coca-Cola over Pepsi is equal to 0.50.
(ii) The rejection point method
Since the z-statistic is less than the critical value, we fail to reject the null hypothesis that the true proportion, p, of cola drinkers that prefer Coca-Cola over Pepsi is equal to 0.50.
Question 3
n1 = 15
n2 = 16
Ho: µ1-µ2 = 0
H1: µ1-µ2 ≠ 0
Sp = sqrt(((15-1)*3.4^2+(16-1)*3.8^2)/(31-2))
= 3.6124
t = 69-72-03.6124*115+116
= -31.298299
= -2.3107
Degrees of freedom = (31-2)
= 29
p-value = 0.01407901
Critical value = 2.04523
(i) The p-value method
Since the p-value is less than 0.05, we reject the null hypothesis that µ1-µ2 = 0 in favor of the alternative hypothesis. Therefore, there is a difference in the average score between the two methods.
(ii) The rejection point method
Since the test statistic is less than the critical value, we reject the null hypothesis that µ1-µ2 = 0 in favor of the alternative hypothesis. Therefore, there is a difference in the average score between the two methods.
Question 4
Old Method
New Method
36
29
55
42
28
30
40
32
62
56
Mean
44.2
37.8
Variance
195.2
130.2
n1 = 5
n2 = 5
Ho: µ1-µ2 = 0
H1: µ1-µ2 > 0
Sp = sqrt(((5-1)*195.2+(5-1)*130.2)/(10-2))
= 12.75539
t = 195.2-130.2-012.75539*25
= 3.222921
Degrees of freedom = (10-2)
= 8
Critical value = 3.355387
Since the test statistic is less than the critical value, we fail to reject the null hypothesis at 5 %. Therefore, the new method does not decrease the average length of the surgical procedure.
Question 5
H0: p=0.25
H1: p<0.25
n = ...

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