H1: at Least One of the Proportions Is Different from the Others

HWK#4

1)Determine (a) the the tastistic, (b) the degrees of freedom, (c) the critical value using a=0.05, and (d) the hypothesis at the a=0.05 level of significance

Outcome A B C D

Observed 45 54 48 53

Expeted 50 50 50 50

H0: PA=PB=PC=PD= ¼

H1: at least one of the proportions is different from the others.

(a) = (…..) Round to three decimal places as needed.

(b)There are (…..) degrees of freedom.

(c)The critical value is (….) Round to three decimal places as needed.

(d)Should the null hypothesis be rejected?

A-Yes because is less than or equal to 0.05

B-No because is greater than 0.05

C-No because 0 is less than or equal to 0.05

D-Yes because 0 is greater than 0.05

2)A study was conducted to determine the effectiveness of a certain treatment. A group of 112 patients were randomly divided into experimental group and a control group. The shows the result for their net improvement. Let the experimental group to be 1 and the control group be group 2.

Experimental Control

n 6151

x 11.2 3.4

s 6.4 6.6

(a)Test whether the experimental group experienced a large mean improvement than the control group at the a = 0.01 level of significance

A-No, because the test statistic is not in the critical region

B-No, because the test statistic is in the critical region

C-Yes, because the test statistic is not in the critical region

D-Yes, because the test statistic is in the critical region

(b)Construct a 95% confidence interval about µ1 - µ2 and interpret the results.

Therefore, the confidence interval is the range from (…..) to (….) (Round to three decimal places as needed)

What is the interpretation of this confidence interval?

A-There is a 95% probability that the difference of the means is in the interval.

B-We are 95% confidence that the difference of the means is in the interval

C-There is a 95% probability that the difference between randomly selected individuals will be in the interval

D-We are 95% confident that the difference between randomly selected individuals will be in the interval.

3)A researcher takes measurement of water clarity at the same location in a lake on the same date during the course of a year and repeats the measurements on the same dates 5 years later. The collected data is given in the table below.

Observation 1 2 3 4 5 6

Date 1/25 3/19 5/30 7/3 9/13 11/7

Initial 62.7 72.9 36.2 61.7 67.2 50.6

After 5 years 66.5 76.2 38.8 69.8 62.9 58.8

(a)Why is it important to take the measurements on the same date?

A-Using the same dates makes it easier to remember to take samples.

B-Using the same dates maximizes the difference in water clarity

C-Those are the same dates that all biologists use to take water clarity samples.

D-Using the same dates makes the second sample dependent on the first.

(b)Is the clarity of the lake improving at the a = 0.05 level of significance? You may want to note that the normal probability plot and boxplot of the data indicate that the differences are approximately normally distributed with no outlier. What is your conclusion regarding H0?

A-Do not reject H0

B-Reject H0

(c)Construct a 95% confidence interval about the population mean different. Compute the difference as (initial depth of clarity) minus (depth of clarity after five years). Interpret your results

The confidence interval is (…..), (…..)(Round to three decimal places as needed.)

Choose the statement which best agrees with your interpretation of your results

A-I’m 5% confident that the means difference lies within the interval

B-I’m 95% confident that the means difference is not zero.

C-I’m 95% confident that the means difference lies within the interval.

D-I’m 95% confidence that the mean difference is 0.

4)The first significant digit in any number must be 1, 2, 3, 4, 5, 6, 7, 8 or 9. It was discovered that first digits do not occurs with equal frequency. Probabilities of occurrence to the first digit in a number are shown in the accompanying table. The probability distribution is now known as Benford’s law. For example, the following distribution represents the first digits in 189 allegedly fraudulent checks written to a bogus company by an employee attempting to embezzle funds from his employer. Complete parts (a) and (b).

Distribution of the first digits

Digit 1 2 3 4 5 6 7 8 9

Probability 0.301 0.176 0.125 0.097 0.079 0.067 0.058 0.051 0.046

First digit allegedly fraudulent checks

First digit 1 2 3 4 5 6 7 8 9

Frequency 36 252820 24 17 15 17 7

(a)Using the level of significance a = 0.05, test whether the first digits in the allegedly fraudulent checks obey Benford’s law.Do the first digits obey the Benford’s law?

Yes

(b)Based on the results of part (a), could one think that the employee is guilty of embezzlement?

Yes

5)The following data represents the muzzle velocity(in feet per second) of shells fired from a 155-mn gun. For each shell two measurements of the velocity were recorded using two different measuring devices resulting in the following data.

Observation 1 2 3 4 5 6

A 794.4790.2 791.3 792.9 790.3 794.9

B 799.6 798.1 787.6 793.5 790.0 792.5

(a)Why are these matched -pairs data?

A-All the measurements came from round fired from the same gun.

B-Two measurements (A and B) are taken on the same rounds

C-The same round was fired In every trial

D-Two measurements (A and B) are taken by the same instrument

(b)Is there a difference in the measurement of the muzzle velocity between device A and B at the a = 0.01 level of significance? Note: A normal probability plot and boxplot of the data indicate that the differences are approximately normal distributed with no outlier. What is your conclusion regarding H0?

A-Reject H0

B-Do not reject H0

(c)Construct a 99% confidence interval about the population mean difference. C computer the difference as device A minus device B. Interpret you results.

The confidence interval is (…….),(…….) (Round to three decimal place as needed).

Choose the statement that best agrees with your interpretation of your results.

A-I’m 1% confident that the mean difference in measurement lies in the interval found above

B-I’m 15% confident that the mean difference in measurement is 0.01

C-I’m 99% confident that the mean difference in measurement lies in the interval found above

D-I’m 15% confident that the mean difference in measurement is 0

6) A physical therapist wanted to know whether the mean step pulse of men was less than the mean step pulse of women. She randomly selected 56 men and 72 women to participate in the study. Each subject was required to step up and down a 6- inch platform. The pulse of each subject was then recorded. The following results were obtained.

Two simple T for men vs Women

N Mean St Dev SE Mean

Man56112.711.21.5

Women72118.614.51.7

99% CI for mu Men - mu Women (- 11.91, 0.11)

T-Test mu Men = mu Women ( vs <)

T= -2.60 P=0.0053 DF= 125

(a)State the null and alternative hypothesis. Which of the following is correct?

A-H0: µ1=µ2; Ha: µ1 > µ2

B-H0: µ1=µ2; Ha: µ1 ≠ µ2

C-H0: µ1=µ2; Ha: µ1 < µ2

(b) Identify the P-value and state the researcher’s conclusion if the level of significance was a =0.001. What is the P-value? P = (……)

State the researcher’s conclusion. Which of the following is correct?

A-Reject H0, there is sufficient evidence to conclude that the mean step pulse of men was less than the mean step pulse of women

B-Reject H0, there is not sufficient evidence to conclude that the mean step pulse of men was less than the mean step pulse of women

C-Fail to reject H0, there is sufficient evidence to conclude that the mean step pulse of men was less than the mean step pulse of women

D-Fail to Reject H0, there is not sufficient evidence to conclude that the mean step pulse of men was less than the mean step pulse of women

(c)What is the 99% confidence interval for the mean difference in pulse rates of men vs women?

(…….),(……..) (Use ascending order. Round to two decimal places as needed)

Interpret the result.

A-99% percent of the times the means are in the confidence interval.

B-We are 99% confident that the mean difference is in the confidence interval.

C-We are 99% confident that the means are in the confidence interval.

D-99% percent of the time the mean difference is in the confidence interval

7) Suppose there is sufficient evidence to reject H0:µ1=µ2=µ3 using a one- way ANOVA. The mean square error from the ANOVA is determined to be 24.8. The sample means are x1 = 3.4, x2 = 10.7, and x3 = 19.5, with n1=n2=n3 = 2. Use Turkey’s test to determine which pair wise means are significantly different using a family wise error rate of a = 0.01.

Based on this data, (Reject/ Do not reject) the hypothesis H0:µ1=µ2, (Do not reject/ Reject) the hypothesis H0:µ1=µ3, and (Reject / Do not reject) the hypothesis H0:µ2=µ3.

8) Researchers want to determine if the psychological profile of healthy children was different that for children suffering from current abdominal pain (RAP) or recurring headaches. A total of 210 children and adolescents were studied and their psychological profiles were graded according to the Child Behavior Checklist 4-18 (CBCL). Children were stratified in two age groups: 4 to 11 years and 12 to 18 years. The results of the study are summarized in the accompanying table.

n Sample mean Sample variance

Control groups 7011.721.5

RAP 709.013.1

Headache7012.48.4

(a)Compute the sample standard deviations for each group.

The sample standard deviation for the control group is Sc = (…….) (Round to the nearest tenth as needed).

Thank sample standard deviation for children suffering from recurrent abdominal pain (RAP) is Sr = (……….) (Round to the nearest tenth as needed)

The sample standard deviation for children suffering from recurring headache is Sh = (……..)

(b)What sampling method was used for each treatment group? Why?

A-Systematic sample because it decreases the sample variabilities

B-Simple random sampling because the researcher wanted to be sure that the distributions of the sample mean are approximately normal

C-Cluster sampling because it eliminates bias from the researcher and subject.

D-Stratified sampling because the researcher wanted to be sure that both children and adolescents were represented in the sample

(c)Use a two simple t-test for independent samples to determine if there is a significant difference in mean CBCL score between the control group and the RAP group (assume that both samples are sample random samples) Use a = 0.05.

State the appropriate conclusion. Choose the correct answer below.

A-There is not enough evidence to indicate that the mean CBCL scores are different for the two groups.

B-There is enough evidence to indicate that the mean CBCL scores are different for the two groups.

(d)It is necessary to check the normality assumption to answer par (c)?

A-Yes

B-No

(e)Use the one way ANOVA procedure with a = 0.05 to determine if the mean CBCL scores are different for the three treatment groups. Choose the correct answer below

A-There is enough evidence to indicate that at least one of the mean score is different

B-There is not enough evidence to indicate that at least one of the mean score is different

(f)Based on your results from part (c) and (e), can you determine if there is a significant difference between the mean score of the RAP group and the headache groups?

A-Yes

B-No

9) Researchers studied the effect that varying level of Ritalin and Adderall had on a child’s ability to follow rules when the child is diagnosed with attention deficit hyperactivity disorder (ADHD). The randomly assigned children to one o the five treatment groups: placebo, 10mg Ritalin, 17.5mg of Ritalin, 7.5mg of Adderall, or 12.5mg of Adderall twice a day. They recorded a score that indicated the child’s ability to follow rules, with a higher score indicating a higher ability to follow rules. The accompanying data are based on their study. Complete parts (a) through (C).

Placebo Ritalin 10mg Ritalin 17.5mg Adderall 7.5mg Adderal 12.5mg

47 98 37 59 80

19 758572115

336077 8281

584445 9291

288389 7374

706962 5591

25 57101 70 62

(a) Test the null hypothesis that the mean score for each treatment is the same at the a = 0.05 level of significance. Note that the requirements for a one-way ANOVA are satisfied. Choose the correct answer below.

A- Do not reject the null hypothesis

B- Reject the null hypothesis

(b)If the null hypothesis is rejected in part (a), use turkey’s test to determine which pairwise means differ using a familywise error rate of a = 0.05. Choose the correct answer below.

A- µp ≠µR10 = µR17.5 =µA7.5 = µA12.5

B- µp = µR10 = µR17.5 =µA7.5 ≠ µA12.5

C-µp ≠ µR10 = µR17.5 ≠µA7.5 = µA12.5

D-µp = µR10 ≠ µR17.5 =µA7.5 = µA12.5

E-The null hypothesis was not rejected.

(c) Draw boxplot of the five treatment levels to support the analytic result obtained in parts (a) and (b). Choose the correct answer below.

A12.5

A7.5

R17.5

R10

-30 60 150

A12.5

A7.5

R17.5

R10

-30 60 150

A12.5

A7.5

R17.5

R10

-30 60 150

10) A stock analyst wondered whether the mean rate of return of financial, energy, and utility stocks differed over the past 5 years. He obtained a simplerandom sample of eight companies from each of the three sectors and obtained the 5 years rates of return shown in the companying table (in percentage). Complete part (a) through (d).

(a) State the null and alternative hypothesis. Choose the correct answer below.

Financial: 10.76; 15.12; 17.21; 5.07; 19.50; 8.21; 10.45; 6.75.

Energy : 12.89; 13.96; 6.33; 11.23; 18.79; 20.73; 9.60; 17.40

Utilities : 11.88; 5.76; 13.67; 9.90; 3.95; 3.44 ; 7.11; 15.70

A- H0 at least one of the means is different and H1: µF = µE = µU

B-H0: µF = µE and H1 the means are different

C-H0:µF = µE = µU and H1: µF < µE < µU

D-H0: µF = µE = µU and H1: at least one of the means is different.

(b) Normal probability plots indicate that the sample data come from normal populations. Are the requirements to use the one- way ANOVA procedure satisfied?

A-No

B-Yes

(c)Are the mean rates of return different at the a = 0.05 level of significance?

State the appropriate conclusion. Choose the correct answer below

A-There is sufficient evidence to support the hypothesis that a least one of the means is different.

B-There is not enough evidence to support the hypothesis that a least one of the means is different.

(d)Draw the boxplot of trhee sectors to support the results obtained in part (c). Choose the correct answer.

ABC

UtilitiesEnergy Utilities

Energy UtilitiesEnergy

Financial Financial Financial

3 13 23 3 13 23 3 13 23

Rate of return% Rate of return % Rate of return %

11) (a) Find the critical value from the studentized range distribution for a =0.01, v 11, and k = 16

(b) Find the critical value from the studentized range distribution for a =0.05, v 12, and k = 17

(d) Find the critical value from the studentized range distribution for H0:µ1=µ2=µ3=µ4=µ5, with n=20 at a = 0.01.

(a) (……..) (Round to three decimal places as needed)

(b) (…….)(Round to three decimal places as needed)

(d) (…….) (Round to three decimal places as needed)

12) Fill in the ANOVA table.

Source of variation Sum of squares Degrees of Freedom Mean Squares F-Test Statistic

Treatment 4224

Error 6476 23

Total

Complete the ANOVA table by filling in the missing values.

Source of variation Sum of squares Degrees of Freedom Mean Squares F-Test Statistic

Treatment 422 4 (…..) (…..)

Error 6476 23 (……)

Total (……..) (………)

(Type the integer or decimal rounded to three decimal places as needed)

13)(a) Find the critical value from the studentized range distribution for a =0.01, v 30, and k = 19

(b) Find the critical value from the studentized range distribution for a =0.01, v 11, and k = 16

(d) Find the critical value from the studentized range distribution for H0:µ1=µ2=µ3=µ4=µ5, with n=23 at a = 0.01.

(a) (……..) (Round to three decimal places as needed)

(b) (…….)(Round to three decimal places as needed)

(d) (…….) (Round to three decimal places as needed)