BUSA 3000 Quantitative Analysis for Business

BUSA 3000 Quantitative Analysis for Business

Homework Assignment 2 (40 points)

Name /Last Name:

IMPORTANT: For each question: First provide the formula and then show your calculations for your answers.

If calculations didn’t provide, a 50% deduction; if formulas didn’t provide a 10% deduction; if both didn’t 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.

The questions in this assignment covers these topics:

Chapter 10: Inference about Means and Proportions with Two Populations

Question 1 & 2 10 points

(25%)

Chapter 11. Inferences about Population Variances

Question 3 & 4 10 points

(25%)

Chapter 12. Tests of Goodness of Fit Independence and Multiple Proportions Question 5 & 6 10 points

(25%)

Chapter 13. Experimental Design and Analysis of Variance Question 7 & 8 10 points

(25%)

You can find similar questions with their answers in the exercise file of each chapter. These files located under each folder of related Chapters.

Please, provide one file as a pdf or doc for 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. Homework assignments will be available for approximately 7 days. I will not accept any excuse or any submissions through email.

Please, provide your answers under each question.

Chapter 10: Inference about Means and Proportions with Two Populations

Question 1 (5 points)

Consider the following hypothesis test.

The following results are for two independent samples taken from the two populations.

a.What is the value of the test statistic? (2 points)

b.What is the p-value? (2 points)

c.With a α=0.5, what is your hypothesis testing conclusion? (1 point)

Question 2 (5 points)

Consider the following data for two independent random samples taken from two normal populations.

a.Compute the two sample means. What is the point estimate of the difference between the two population means? (1 point)

b.Compute the two sample standard deviations. (2 points)

c. What is the degrees of freedom for the t distribution? (1 point)

d. What is the 90% confidence interval estimate of the difference between the two population means? (1 point)

Chapter 11: Inferences about Population Variances

Question 3 (5 points)

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 α= .05 to test whether the population variance specification is being violate (use p value approach)

H0: 2 .0004

Ha: 2 .0004 Research hypothesis

Question 4 (5 points)

Two new assembly methods are tested and the variances in assembly times are reported. Use α=.10 and test for equality of the two population variances.

The hypothesis

H0: 1 2 = 2 2

Ha: 1 2 ≠ 2 2

Computing the value of the test statistic

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 – 1 = 24

0.05 = 2 x 0.025 < P < 2 x 0.05 = 0.10

P < 0.10, Therefore, reject Ho.

Chapter 12: Tests of Goodness of Fit Independence and Multiple Proportions

Question 5 (5 points)

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

H0: pA =.pB = pC = pD = .25

A sample of size 300 yielded the following results.

A:85 B:95 C:50 D:70

Use α=.05 to determine whether H0 should be rejected. What is the p-value?

Expected frequencies: (1 point)

Actual frequencies: (0.5 point)

(1.5 points)

degrees of freedom: k – 1 = (0.5 point)

Using the table with df = ,= …………. shows the p-value is…….. than ……….. (0.5 point)

Conclusion:(0.5 point)

p-value ………….05, ………….H0

Complete this sentence based on your conclusion: (0.5 point)

The population proportions ……………. the same.

Question 6 (5 points)

The following table contains observed frequencies for a sample of 240. Test for independence of the row and column variables using a α=.05.

H0: The column variable is independent of the row variable

Ha: The column variable is not independent on the row variable

Observed Frequencies (fij) (0.5 point)

  A B C Total

P Q R Total Expected Frequencies (eij) (1 point)

  A B C Total

P Q R Total Chi–Square Calculations (fij – eij)2 / eij (1.5 points)

  A B C Total

P Q R =

Degrees of freedom = (r – 1)(c – 1) = (0.5 point)

Using the table with df = ………,=……………….. shows the p–value is ………. than ………… (0.5 point)

Conclusion: (0.5 point)

p–value ……….. .05, ……………. H0.

Complete this sentence based on your conclusion: (0.5 point)

The column variable …………… independent of the row variable.

Chapter 13: Experimental Design and Analysis of Variance

Question 7 (5 points)

The following data are from a completely randomized design. (Data file can be found in the D2L.)

a.Compute the sum of squares between treatments. (.5 point)

x = 119 + 107 + 100/3 = 107.9

SSTr = 8*(119 – 107.9) ^2 + 10*(107 – 107.9) ^2 + 10*(100 – 107.9) ^2 = 1617.86

dfBG = 3 -1

= 2

The mean square between groups is,

MSBG = SSdf=1617.86/2 = 808.93

b.Compute the mean square between treatments. (.5 point)

MSBG = SSdfMSTr = 1617.86/2 = 808.929

c.Compute the sum of squares due to error. (1 point)

SSE = (8 – 1) *146.86 + (10 – 1) *96.44 + (10 – 1) *173.78 = 3460

d.Compute the mean square due to error. (1 point)

3460/ (10 + 10 + 8 – 2) = 138.4

e. Set up the ANOVA table for this problem. Use the Excel Single Factor ANOVA test and provide screenshot: (1 point)

ANOVA table Source SS    df MS F    p-value

Treatment 1,617.86 2 808.929 5.84 .0083

Error 3,460.00 25 138.400 Total 5,077.86 27 f.At α= .05, is there a significant difference between the treatment means? (1 points)

Since the p-value (0.0083) is less than the significance level (0.05), we can reject the null hypothesis.

Therefore, we can conclude that there is a significant difference between treatment means.

Question 8 (5 points)

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?

a. Complete the following ANOVA table. (3 points)

b. What hypotheses are implied in this problem? (1 point)

The p-value is 0.05

Consider Null and Alternative hypothesis.

Null hypothesis, all treatment means are equal.

Alternative hypothesis, all treatment means are not equal.

Hence, conclude that all treatment means are not equal

All treatment means are not equal.

c. At the α= .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 < 0.01

p < 0.05

There is sufficient evidence to reject the claim of equal population means.

Top of Form

Source

of Variation Sum

of Squares Degrees

of Freedom Mean

Square F p-value

Treatments 1200 2 600 44 < 0.01

Error 600 44 13.6364 Total 1800 46