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Question Details:
#3718

Assignment 1

The number of attempts available for each question is noted beside the question. If you are having trouble figuring out your error, you should consult the textbook, or ask a fellow student, one of the TA’s or your professor for help. There are also other resources at your disposal, such as the Engineering Drop in Centre and the Mathematics Continuous Tutorials. Don’t spend a lot of time guessing – it’s not very efficient or effective. Make sure to give lots of significant digits for (floating point) numerical answers. For most problems when entering numerical answers, you can if you wish enter elementary expressions such as 2^3 instead of 8, sin(3 pi=2)instead of -1, e^(ln(2)) instead of 2, (2+tan(3)) (4sin(5))^67=8 instead of 27620.3413, etc.

1. (1 pt) An agricultural field trial compares the yield of two varieties of corn. The researchers divide in half each of 19 fields of land in different locations and plant each corn variety in one

half of each plot. After harvest, the yields are compared in bushels per acre at each location. The 19 differences (Variety A - Variety B) give ¯ x = 2:27 and s = 4:49. Does this sample provide evidence that Variety A had a higher yield than Variety B?

(a) State the null and alternative hypotheses: (Type ”mu” for the symbol μ , e.g. mu > 1 for the mean is greater than 1, mu < 1 for the mean is less than 1, mu not = 1 for the mean is not equal to 1)( Hint: Use the only equal sign in the null hypothesis)

H0 :

HA :

(b) Find the value of the test statistic. Use two decimal places.

(c) Answer the question: Does this sample provide evidence that Variety A had a higher yield than Variety B? (Use a 5% level of significance) (Type: Yes or No)

Answer(s) submitted:

2. (1 pt) Credit scores are used by banks, financial institutions, and retailers to determine one’s trustworthiness, and their ability to pay their credit card, when he/she is given a credit card. The higher the credit score, the better the credit (and the more financially trustworthy) a consumer is. In light of the current economic downturn, an economist claims that the credit score of Canadians between the ages of 25 to 40 has dropped. Prior to the current economic downturn, the mean credit score for Canadians between the ages of 25 to 40 was 673. A random sample of n = 20 Canadians between the ages of 25 to 40 was taken, the credit score of each was determined using the a certain credit bureau. The raw data is given below. 656;715;615;670;553;683;678;610;644;719;673;651;584;669;652;658;651;591;676;667

(a) Choose the correct statistical hypotheses.

A. H0 : μ = 673; HA : μ 6= 673

B. H0 : X > 673; HA : X < 673

C. H0 : X = 673; HA : X < 673

D. H0 : μ > 673; HA : μ < 673

E. H0 : X = 673; HA : X 6= 673

F. H0 : μ = 673 HA : μ < 673

(b) Determine the value of the test statistic for this test, use

two decimals in your answer.

Test Statistic =

(c) Determine the P-value for this test, to three decimal places.

P =

(d) Based on the above calculations, we should ? the null

hypothesis. Use alpha = 0.05

Answer(s) submitted:

3. (1 pt) Is the number of American-born hockey players

playing in the NHL on the rise? In the 2010-2011 season,

24born in the United States.

A random sample of n = 110 NHL hockey players found that

28 were born in the United States.Does this sample suggest that

the percentage of NHL hockey players that are American-born

is increasing?

(a) Choose the correct statistical hypotheses.

A. H0 : p > 0:24; HA : p 0:24

B. H0 : p = 0:24; HA : p 6= 0:24

C. H0 : bp < 0:24; HA : bp 0:24

D. H0 : p = 0:24 HA : p > 0:24

E. H0 : bp = 0:24; HA : bp > 0:24

1

(b) Determine the value of the test statistic for this test, to

two decimal places.

Test Statistic =

(c) Determine the P-value for this test. Use three decimal places.

P =

(d) Based on the above calculations, we should ? the null

hypothesis. Use alpha = 0.05

Answer(s) submitted:

4. (1 pt) According to data from the Tobacco Institute Testing

Laboratory, a certain brand of cigarette contains an average

of 1.4 milligrams of nicotine. An advocacy group questions

this figure, and commissions an independent test to see if the

the mean nicotine content is higher than the industry laboratory

claims.

The test involved randomly selecting n = 15 cigarettes, measuring

the nicotine content (in milligrams) of each cigarette. The

data is given below.

1:7;1:6;1:8;2:0;1:4;1:4;1:9;1:6;1:3;1:5;1:2;1:4;1:7;1:2;1:5

Assume that the nicotine content in each cigarette varies,

which can be modeled by the Normal distribution.

(a) Choose the correct statistical hypotheses.

A. H0 : X = 1:4; HA : X < 1:4

B. H0 : μ = 1:4 HA : μ > 1:4

C. H0 : X > 1:4; HA : X < 1:4

D. H0 : X = 1:4; HA : X 6= 1:4

E. H0 : μ = 1:4; HA : μ 6= 1:4

F. H0 : μ > 1:4; HA : μ < 1:4

(b) Determine the value of the test statistic for this test, use

two decimals in your answer.

Test Statistic =

(c) Determine the P-value for this test, to three decimal places.

P =

(d) Based on the above calculations, we should ? the null

hypothesis. Use alpha = 0.05

Answer(s) submitted:

5. (1 pt) A statistical investigation is conducted to see if the

mean of a population of values is equal to 350, or

H0 : μ = 350 HA : μ 6= 350:

A random sample of n = 100 values is taken from this population

of values, producing a test statistic of Tcalc = 0:75 and a

P-value of 0:4550:

Choose the correct interpretation of the P-value.

A. If the null hypothesis is true, the chance of another

sample producing stronger evidence against the null hypothesis

is 0.4550.

B. If the null hypothesis is true, the probability of rejecting

the null hypothesis is 0.4550.

C. From this sample, the null hypothesis is to be rejected.

D. From this sample, the null hypothesis is not to be

rejected.

Answer(s) submitted:

6. (1 pt) A 2007 Carnegie Mellon University study reported

that online shoppers were willing to pay, on average, more than

an extra $0.60 on a $15 purchase in order to have better online

privacy protection.

A sample of n = 22 online shoppers was taken, and each was

asked “how much extra would you pay, on a $15 purchase, for

better online privacy protection?” The data is given below, in

$’s.

0:79;0:41;0:67;0:67;0:83;0:76;0:55;0:92;0:61;0:57;0:54;1:25;0:70;0:85;(a) Choose the correct statistical hypotheses.

A. H0 : μ > 0:60; HA : μ < 0:60

B. H0 : X = 0:60; HA : X < 0:60

C. H0 : μ = 0:60 HA : μ > 0:60

D. H0 : X = 0:60; HA : X > 0:60

E. H0 : μ = 0:60; HA : μ 6= 0:60

F. H0 : μ > 0:60 HA : μ = 0:60

2

(b) Determine the value of the test statistic for this test, using

two decimals in your answer.

Test Statistic =

(c) Determine the P-value for this test, enter your answer to

three decimals.

P =

(d) Based on the above calculations, we should ? the null

hypothesis. Use alpha = 0.05

Answer(s) submitted:

(incorrect)

7. (1 pt)

In the judicial system, the defense attorney argues for the null

hypothesis that the defendant is innocent. In general, what

would be the result if judges instructed juries to ...

a) never make a type I error, the jury would then be forced

to ?

b) never make a type II error, the jury would then be forced

to ?

Answer(s) submitted:

8. (1 pt) A poll of university students in Canada found that

one-quarter of all students completing an undergraduate program

have 2 or more credit cards.

A random sample of n = 400 university students who recently

completed an undergraduate program found that 112 had 2 or

more credit cards. Does this sample support the one-quarter parameter?

(a) Choose the null and alternative hypotheses

A. H0 : p = 0:25; HA : p < 0:25

B. H0 : p = 0:25 HA : p 6= 0:25

C. H0 : bp = 0:25; HA : bp 6= 0:25

D. H0 : p = 0:25; HA : p > 0:25

E. H0 : bp = 0:25; HA : bp < 0:25

(b) Determine the value of the test statistic for this test, to

two decimal places.

Test Statistic =

(c) Determine the P-value for this test. Use three decimal places.

P =

(d) Based on the above calculations, we should ? the null

hypothesis. Use alpha = 0.05

Answer(s) submitted:

9. (1 pt) A statistical investigation is conducted to see if the

mean of a population of values is equal to 200, or

H0 : μ = 200 HA : μ 6= 200

A random sample of n = 60 values is taken from this population

of values, from which the mean X and the standard deviation S

were computed. From these, the test statistic was found to equal

0:75.

(a) Find the P-value. Use four decimals.

P =

(b) Choose the correct interpretation of the P-value found in

(a).

A. The probability computed in (a) is the chance of another

sample producing stronger evidence against the

null hypothesis, assuming that the null hypothesis is

true.

B. The probability in (a) is the probability of rejecting

the null hypothesis.

C. The probability in (a) is the probability of concluding

the null hypothesis to be true.

D. The probability computed in (a) is the probability

of rejecting the null hypothesis, assuming that the null

hypothesis is true.

Answer(s) submitted:

10. (1 pt) A new chemotherapy regime for the treatment of

ovarian cancer consisting of a combination of gemcitabine and

cisplatin, or GEMCIS, is to be treated on women with Stage III

ovarian cancer. The null hypothesis is that the GEMCIS treatment

is not more effective than the current treatment protocol,

one that only uses cisplatin. In the context of a hypothesis test,

(a) select the correct explaination of a Type I error.

A. One would conclude that GEMCIS is not a more

effective treatment of ovarian cancer than the current

treatment protocol, when in fact GEMCIS is not effective

than the current treatment protocol.

B. One would conclude that GEMCIS is a more effective

treatment of ovarian cancer than the current treatment

protocol, when in fact GEMCIS is more effective

than the current treatment protocol.

C. One would conclude that GEMCIS is not a more

effective treatment of ovarian cancer than the current

treatment protocol, when in fact GEMCIS is more effective

than the current treatment protocol.

D. One would conclude that GEMCIS is a more effective

treatment of ovarian cancer than the current treatment

protocol, when in fact GEMCIS is not more effective

than the current treatment protocol.

(b) select the correct explaination of a Type II error.

A. One would conclude that GEMCIS is not a more

effective treatment of ovarian cancer than the current

treatment protocol, when in fact GEMCIS is not more

effective than the current treatment protocol.

B. One would conclude that GEMCIS is a more effective

treatment of ovarian cancer than the current treatment

protocol, when in fact GEMCIS is more effective

than the current treatment protocol.

C. One would conclude that GEMCIS is a more effective

treatment of ovarian cancer than the current treatment

protocol, when in fact GEMCIS is more effective

than the current treatment protocol.

D. One would conclude that GEMCIS is not a more

effective treatment of ovarian cancer than the current

treatment protocol, when in fact GEMCIS is more effective

than the current treatment protocol.

Answer(s) submitted:

11. (1 pt) Hourly wages at Manufacturing Plant A very from

worker to worker, with an average hourly wage of $14.00, and

a standard deviation of $2.50 per hour, or sPlantA = 2:50. In

an attempt to attract new employees, a newly constructed Manufacturing

Plant B claims that its employees will be paid, on

average, more than than employees at Manufacturing Plant A.

Assume that the standard deviation in hourly wages at Plant B

is the same as at Plant A, or sPlantB = 2:50.

(a) Choose the correct statistical hypotheses.

A. H0 : μPlantB 14:00 HA : μPlantB < 14:00

B. H0 : μPlantB = 14:00 HA : μPlantB 6= 14:00

C. H0 : XPlantB = 14:00 HA : XPlantB > 14:00

D. H0 : XPlantB > 14:00 HA : XPlantB < 14:00

E. H0 : μPlantB = 14:00 HA : μPlantB > 14:00

F. H0 : μPlantB > 14:00 HA : μPlantB = 14:00

(b) Statistical testing of the null hypothesis is to be carried

out by randomly selecting 40 employees at Plant B. If the

mean/average hourly wage of these 40 employees is greater

than $14.77, then there is enough statistical evidence to indicate

that the mean hourly wage of employees at Plant B is greater

than the mean hourly wage of employees at Plant A.

What level of a was used here? Enter your answer to three

decimal places.

(c) The average of the sample of n = 40 workers was found

to be X = 14:33. What decision can you make from this sample?

A. The sample size is too small to conduct a statistical

test, so a decision cannot be made

B. Employees at Plant B earn do not more on average

compared to employees at Plant A.

C. Employees at Plant B do earn more on average compared

to employees at Plant A

D. The hourly wages at Plant B need to be Normally

distributed to do the test

(d) If the average hourly wage of all workers at Plant B is

$14.50, what is the probability of concluding using the decision

criteria outlined in part (b) that the mean hourly wage of all employees

at Plant B is not greater than the mean hourly wage of

all employees at Plant A? Enter your answer to three decimals.

Answer =

Answer(s) submitted:

12. (1 pt) The editor of a magazine knows that 40

(a) Choose the correct statistical hypotheses.

A. H0 : bp = 0:40 HA : bp < 0:40

B. H0 : p = 0:40 HA : p 6= 0:40

C. H0 : p = 0:40 HA : p > 0:40

D. H0 : bp = 0:40 HA : bp 6= 0:40

E. H0 : bp = 0:40 HA : bp > 0:40

F. H0 : p = 0:40 HA : p < 0:40

(b) The null hypothesis is to be tested from a random sample

of n = 200 magazine subscribers whose subscriptions were

to be renewed in the past six months. If statistical testing is to

be carried out at P(Type I Error) = 0:05, what is the maximum

number of subscribers (or minimum) out of 200 who need to

have renewed their subscription in order to determine that the

renewal offer does decrease the percentage of subscribers who

do not renew. Enter your answer to the nearest integer.

Number of subscribers needed to not renew in order for percentage

to decrease is =

(c) The sample of 200 subscribers who were offered the renewal

offer was taken, of which 74 renewed their magazine

subscription in the past six months. The P-value of this result

was 0:1930. What type of error could be made from this

sample?

A. Type III Error

B. Type I Error

C. Type II Error

D. Power of a Test

(d) How powerful is the statistical test outlined in (b) if the

percentage of all subscribers who do not renew their subscription,

p, drops by 5 percentage points ? Enter your answer to

three decimals.

Power =

Answer(s) submitted:

(incorrect)

13. (1 pt) The owner of a dry cleaning store believes that the

mean amount a customer spends on a dry cleaning order exceeds

$22.00. A statistician has said to him that he can be statistically

sure that his belief is true if the mean amount of 100 randomly

chosen customer bills is greater than $22.49, or X > 22:49. It is

assumed that the standard deviation in the bill amounts is $2.50

(s = 2:50).

(a) Choose the correct statistical hypotheses.

A. H0 : X = 22; HA : X > 22

B. H0 : X > 22:49; HA : X 22:49

C. H0 : μ > 22; HA : μ < 22

D. H0 : X = 22; HA : X 6= 22

E. H0 : X = 22:49; HA : X > 22:49

F. H0 : μ = 22; HA : μ 6= 22

G. H0 : μ = 22 HA : μ > 22

(b) What is the probability of making a Type I error, using

the statistician’s criterion? Use three decimals in your answer.

P(Type I) =

(c) Unknown to anyone, suppose the mean amount spent by

all his customers is $21.75. Find the probability that the owner

will conclude that his original belief is correct. Use three decimals

in your answer.

Answer =

(d) Suppose the statistician decides to change the sample size to

n = 200 and regulate P(Type I) = 0:05. For what values of X

should the null hypothesis in (a) be rejected?

A. Reject the null hypothesis if X > 22:49

B. Reject the null hypothesis if μ > 22:50

C. Reject the null hypothesis if X > 22:00

D. Reject the null hypothesis if X > 22:50

E. Reject the null hypothesis if μ > 22:29

F. Reject the null hypothesis if X > 22:29

Answer(s) submitted:

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Solution Details:
#3707

Please find answers in the attached file. One item ...

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Solution Details:
#3707

P-value of this test = 2P (T18 > 2.20) = 0.041
Since P-value < α = 0.05, reject the null hypothesis.
Therefore, this sample provide evidence t...

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