BUSA 3000 Quantitative Analysis for Business<\/p>\n
Homework Assignment 2 (40 points)<\/p>\n
Name \/Last Name: <\/p>\n
IMPORTANT: For each question: First provide the formula and then show your calculations for your answers. <\/p>\n
If calculations didn\u2019t provide, a 50% deduction; if formulas didn\u2019t provide a 10% deduction; if both didn\u2019t provide, a 70% deduction will be conducted. If you are going to provide work and have mistakes still you can get partial points for your work.<\/p>\n
The questions in this assignment covers these topics:<\/p>\n
Chapter 10: Inference about Means and Proportions with Two Populations <\/p>\n
Question 1 & 2 10 points<\/p>\n
(25%)<\/p>\n
Chapter 11. Inferences about Population Variances <\/p>\n
Question 3 & 4 10 points<\/p>\n
(25%)<\/p>\n
Chapter 12. Tests of Goodness of Fit Independence and Multiple Proportions Question 5 & 6 10 points<\/p>\n
(25%)<\/p>\n
Chapter 13. Experimental Design and Analysis of Variance Question 7 & 8 10 points<\/p>\n
(25%)<\/p>\n
You can find similar questions with their answers in the exercise file of each chapter. These files located under each folder of related Chapters.<\/p>\n
Please, provide\u00a0one file as a pdf or doc\u00a0for your HW assignments. So, if you are writing by hand, you have to merge your photos in a pdf file. I will not accept separate pictures, photos. You have to arrange your assignment and submit in a professional manner.\u00a0Homework assignments will be available for approximately 7 days.\u00a0I will not accept any excuse or any submissions through email.<\/p>\n
Please, provide your answers under each question.<\/p>\n
Chapter 10: Inference about Means and Proportions with Two Populations<\/p>\n
Question 1 (5 points)<\/p>\n
Consider the following hypothesis test.<\/p>\n
The following results are for two independent samples taken from the two populations.<\/p>\n
a.What is the value of the test statistic? (2 points)<\/p>\n
b.What is the p-value? (2 points)<\/p>\n
c.With a \u03b1=0.5, what is your hypothesis testing conclusion? (1 point)<\/p>\n
Question 2 (5 points)<\/p>\n
Consider the following data for two independent random samples taken from two normal populations.<\/p>\n
a.Compute the two sample means. What is the point estimate of the difference between the two population means? (1 point)<\/p>\n
b.Compute the two sample standard deviations. (2 points)<\/p>\n
c. What is the degrees of freedom for the t distribution? (1 point) <\/p>\n
d. What is the 90% confidence interval estimate of the difference between the two population means? (1 point)<\/p>\n
Chapter 11: Inferences about Population Variances<\/p>\n
Question 3 (5 points)<\/p>\n
An automotive part must be machined to close tolerances to be acceptable to customers. Production specifications call for a maximum variance in the lengths of the parts of .0004. Suppose the sample variance for 30 parts turns out to be s2= .0005. Use a \u03b1= .05 to test whether the population variance specification is being violate (use p value approach) <\/p>\n
H0: 2 \uf0a3\uf020.0004<\/p>\n
Ha: 2 \uf03e\uf020.0004 Research hypothesis<\/p>\n
Question 4 (5 points)<\/p>\n
Two new assembly methods are tested and the variances in assembly times are reported. Use \u03b1=.10 and test for equality of the two population variances.<\/p>\n
The hypothesis<\/p>\n
H0: 1 2 = 2 2<\/p>\n
Ha: 1 2 \u2260 2 2<\/p>\n
Computing the value of the test statistic<\/p>\n
F = s12s22=2512=2.083The p-value is the probability of obtaining the value of the test statistic, or a value more extreme. The p-value is the number or interval in the column title of Table 4 containing the F-value with df n = 31-1 = 30 and df d = 25 \u2013 1 = 24<\/p>\n
0.05 = 2 x 0.025 < P < 2 x 0.05 = 0.10<\/p>\n
P < 0.10, Therefore, reject Ho.<\/p>\n
Chapter 12: Tests of Goodness of Fit Independence and Multiple Proportions<\/p>\n
Question 5 (5 points)<\/p>\n
Suppose we have a multinomial population with four categories: A, B, C, and D. The null hypothesis is that the proportion of items is the same in every category. The null hypothesis is<\/p>\n
H0: pA =.pB = pC = pD = .25<\/p>\n
A sample of size 300 yielded the following results.<\/p>\n
A:85 B:95 C:50 D:70 <\/p>\n
Use \u03b1=.05 to determine whether H0 should be rejected. What is the p-value?<\/p>\n
Expected frequencies: (1 point)<\/p>\n
Actual frequencies: (0.5 point)<\/p>\n
(1.5 points)<\/p>\n
degrees of freedom: k – 1 = (0.5 point)<\/p>\n
Using the table with df = ,= \u2026\u2026\u2026\u2026. shows the p-value is\u2026\u2026.. than \u2026\u2026\u2026.. (0.5 point) <\/p>\n
Conclusion:(0.5 point)<\/p>\n
p-value \u2026\u2026\u2026\u2026.05, \u2026\u2026\u2026\u2026.H0<\/p>\n
Complete this sentence based on your conclusion: (0.5 point)<\/p>\n
The population proportions \u2026\u2026\u2026\u2026\u2026. the same.<\/p>\n
Question 6 (5 points)<\/p>\n
The following table contains observed frequencies for a sample of 240. Test for independence of the row and column variables using a \u03b1=.05.<\/p>\n
H0: The column variable is independent of the row variable<\/p>\n
Ha: The column variable is not independent on the row variable<\/p>\n
Observed Frequencies (fij) (0.5 point)<\/p>\n
\u00a0 A B C Total<\/p>\n
P Q R Total Expected Frequencies (eij) (1 point)<\/p>\n
\u00a0 A B C Total<\/p>\n
P Q R Total Chi\u2013Square Calculations (fij \u2013 eij)2 \/ eij (1.5 points)<\/p>\n
\u00a0 A B C Total<\/p>\n
P Q R = <\/p>\n
Degrees of freedom = (r \u2013 1)(c \u2013 1) = (0.5 point)<\/p>\n
Using the table with df = \u2026\u2026\u2026,=\u2026\u2026\u2026\u2026\u2026\u2026.. shows the p\u2013value is \u2026\u2026\u2026. than \u2026\u2026\u2026… (0.5 point)<\/p>\n
Conclusion: (0.5 point)<\/p>\n
p\u2013value \u2026\u2026\u2026.. .05, \u2026\u2026\u2026\u2026\u2026. H0. <\/p>\n
Complete this sentence based on your conclusion: (0.5 point)<\/p>\n
The column variable \u2026\u2026\u2026\u2026\u2026 independent of the row variable.<\/p>\n
Chapter 13: Experimental Design and Analysis of Variance<\/p>\n
Question 7 (5 points)<\/p>\n
The following data are from a completely randomized design. (Data file can be found in the D2L.)<\/p>\n
a.Compute the sum of squares between treatments. (.5 point)<\/p>\n
x = 119 + 107 + 100\/3 = 107.9<\/p>\n
SSTr = 8*(119 – 107.9) ^2 + 10*(107 – 107.9) ^2 + 10*(100 – 107.9) ^2 = 1617.86<\/p>\n
dfBG = 3 -1 <\/p>\n
= 2<\/p>\n
The mean square between groups is,<\/p>\n
MSBG = SSdf=1617.86\/2 = 808.93<\/p>\n
b.Compute the mean square between treatments. (.5 point)<\/p>\n
MSBG = SSdfMSTr = 1617.86\/2 = 808.929<\/p>\n
c.Compute the sum of squares due to error. (1 point)<\/p>\n
SSE = (8 – 1) *146.86 + (10 – 1) *96.44 + (10 – 1) *173.78 = 3460<\/p>\n
d.Compute the mean square due to error. (1 point)<\/p>\n
3460\/ (10 + 10 + 8 – 2) = 138.4<\/p>\n
e. Set up the ANOVA table for this problem. Use the Excel Single Factor ANOVA test and provide screenshot: (1 point)<\/p>\n
ANOVA table Source SS \u00a0\u00a0 df MS F \u00a0\u00a0 p-value<\/p>\n
Treatment 1,617.86 2 808.929 5.84 .0083<\/p>\n
Error 3,460.00 25 138.400 Total 5,077.86 27 f.At \u03b1= .05, is there a significant difference between the treatment means? (1 points)<\/p>\n
Since the p-value (0.0083) is less than the significance level (0.05), we can reject the null hypothesis.<\/p>\n
Therefore, we can conclude that there is a significant difference between treatment means.<\/p>\n
Question 8 (5 points)<\/p>\n
In a completely randomized design, 12 experimental units were used for the first treatment, 15 for the second treatment, and 20 for the third treatment. Complete the following analysis of variance. At a .05 level of significance, is there a significant difference between the treatments?<\/p>\n
a. Complete the following ANOVA table. (3 points)<\/p>\n
b. What hypotheses are implied in this problem? (1 point)<\/p>\n
The\u00a0p-value\u00a0is 0.05<\/p>\n
Consider Null and Alternative hypothesis.<\/p>\n
Null hypothesis,\u00a0all treatment means are equal.<\/p>\n
Alternative hypothesis,\u00a0all treatment means are not equal.<\/p>\n
Hence, conclude that all treatment means are not equal<\/p>\n
All treatment means are not equal.<\/p>\n
c. At the \u03b1= .05 level of significance, can we reject the null hypothesis in part (b)? Is there a significant difference between the treatments? Explain.(1 point) <\/p>\n
p < 0.01<\/p>\n
p < 0.05<\/p>\n
There is sufficient evidence to reject the claim of equal population means.<\/p>\n
Top of Form<\/p>\n
Source<\/p>\n
of Variation Sum<\/p>\n
of Squares Degrees<\/p>\n
of Freedom Mean<\/p>\n
Square F p-value<\/p>\n
Treatments 1200 2 600 44 < 0.01<\/p>\n
Error 600 44 13.6364 Total 1800 46<\/p>\n","protected":false},"excerpt":{"rendered":"
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