STATS 250 HW #4

Assignment Stats 250 W11 HW 4

Graded Point Total: 29 out of 30

a.

1 out of 1 points

Let p1 represent the population proportion for all such boys that regularly go on-line. Similarly, p2

represents the population of all such girls that do so. Assuming that each sample is a random sample

from the populations of interest and that the two samples are independent, verify the remaining conditions

necessary for computing a confidence interval for p1 − p2.

Your Answer:

n1 1, n1(1- 1), n2 2, n2(1- 2) should be preferably at least 10.

1 = 156/236 = .6610

n1 1 = .6601 * 236 = 156

n1(1- 1) = 236*(1-.6610) = 80

2 = 152/272 = .5588

n2 2 = 272* .5588 = 152

n2(1- 2) = 272*(1-.5588) = 120

n1 1, n1(1- 1), n2 2, n2(1- 2) are all greater than 10.

Solution:

The sample sizes are sufficiently large enough. For a confidence interval, we need to have that all

of the quantities n1 1 =156, n1(1 - 1) =80, n2 2 =152, and n2(1 - 2) =120 be at least 10. And

we do!

b.

3 out of 3 points

Construct a 95% confidence interval for the difference in these two population proportions, namely p1 − p2.

Your Answer:

Standard Error: s e ( 1 2) = sqrt (( 1*(1 1)/n1)+( 2*(1 2)/ n2))

Required Assignment Stats 250 W11 HW 4

Question 1

(Chapter 10)

Several years ago, Newsweek conducted a survey of 508 teenagers (age 12-17) and asked

questions regarding their thoughts on technology and its impact on their lives. Of the 236 boys

surveyed, 156 stated they regularly go on-line. Of the 272 girls surveyed, 152 stated they

regularly go on-line.

4/17/2011 www.lecturebook.com/homeworks/sho…

lecturebook.com/homeworks/show/39 1/12

Standard Error: s.e.( 1 - 2) = sqrt (( 1*(1- 1)/n1)+( 2*(1- 2)/ n2))

=sqrt((.6610*(1-.6610)/236)+(.5588*(1-.5588)/272))

= .0431

Confidence interval:

=( 1 - 2) +/- z*s.e.( 1 - 2)

=(.6610 - .5588) +/- 1.96 * .0431

=0.1022 +/- 0.084476

=(0.017724, 0.186676)

Solution:

So the 95% confidence interval for the difference in population proportions (boys less girls) that

go on-line regularly is: (0.0178, 0.1866) or 1.78% to 18.66%.

a.

1 out of 1 points

Provide an estimate of p1, including an appropriate symbol for your answer.

Your Answer:

1 = 57/100 = .57

Solution:

Estimate of p1 is 1 = 57/100 = 0.57

b. A 95% confidence interval for the difference in population rates, p1 – p2, was calculated to be (-0.05, 0.23).

In order for this confidence interval to be valid, certain conditions must apply. The problem states that the

samples are independent random samples. Verify the remaining condition(s).

Question 2

(Chapter 10)

High School Sports The Department of Education wants to estimate the difference in rates at

which high school boys and girls participate in school sponsored sports. Let p1 represent the

population proportion of all high school girls that participate in a school sport and let p2

represent the population proportion of all high school boys that participate in a school sport.

Independent random samples were collected and the following results were obtained.

Participate in Sports?

Gender Yes No Total

1 = Girls 57 43 100

2 = Boys 48 52 100



1 out of 1 points

Your Answer:

The remaining conditions are that n1 1, n1(1- 1), n2 2, n2(1- 2) should be preferably at least

10.

1 = 43/100 = .57

n1 1 = .57 * 100 = 57

n1(1- 1) = 100*(1-.57) = 43

2 = 48/100 = .48

n2 2 = 100* .48 = 48

n2(1- 2) = 100*(1-.48) = 52

n1 1, n1(1- 1), n2 2, n2(1- 2) are all greater than 10.

Solution:

In each sample, we need to see at least 10 yes responses and at least 10 no responses, so the

verification is: All of the counts 57, 43, 48, and 52 are at least 10.

c.

0.5 out of 0.5points

True or False? There is a 95% chance that p1 – p2 is between -0.05 and 0.23.

1. True

2. False

Solution: 2. False

d.


True or False? With repeated samples of the same sizes, we’d expect p1 – p2 to fall between -0.05 and

0.23 roughly 95% of the time.

1. True

2. False

Solution: 2. False

e. True or False? With repeated samples of the same sizes, we’d expect 95% of the resulting intervals to

contain p1 – p2.

1. True

2. False




Solution: 1. True

f.


True or False? With 95% confidence, we would estimate the population proportion of all high school girls

that participate in a school sport to be anywhere from 5% below to 23% above the corresponding

proportion for all high school boys.

1. True

2. False

Solution: 1. True

g.


Compared to the 95% interval of (-0.05, 0.23), a 90% confidence interval would be … Select one.

1. Wider

2. Narrower

3. Can't Decide

Solution: 2. Narrower

h.

1 out of 1 points

Suppose you want to test the hypothesis H0: p1 = p2 versus Ha: p1 ≠ p2 at the 5% level. Based on the

confidence interval (-.05, .23), what would be your decision? Select one.

1. Reject H0

2. Fail to reject H0

3. Can't Decide

Solution: 2. Fail to reject H0



a.

2 out of 2 points

Before creating the 95% confidence interval to estimate the difference p1 - p2, where 1 = Treatment X and 2

= the Standard Treatment; it would be important to clearly understand an appropriate interpretation of

these symbols. In general, p is a population proportion, but define the symbols further in the context of this

real-world problem. [Fill in] Let p1 = ____________ and p2 = ____________.

Your Answer:

p1 = the population proportion of patients who recieved treatment x and was cured after 6 months.

p2 = the population proportion of patients who recieved the standard treatment and was cured

after 6 months.

Solution:

Let p1 = the population proportion of all patients cured after receiving Treatment X,

and p2 = the population proportion of all patients cured after receiving Standard Treatment.

b.

3 out of 3 points

Construct a 95% confidence interval for the difference p1 - p2.

95% Confidence Interval: ( _________________ , _________________ )

Your Answer:

Standard Error: s.e.( 1 - 2) = sqrt (( 1*(1- 1)/n1)+( 2*(1- 2)/ n2))

=sqrt((.49*(1-.49)/200)+(.38*(1-.38)/100))

= 0.0600

Confidence interval:

=( 1 - 2) +/- z*s.e.( 1 - 2)

=(.49 - .38) +/- 1.96 * .0600

=0.11 +/- 0.1176

Question 3

(Chapter 12)

Comparing Treatment X and a Standard Treatment — A researcher wishes to compare the

cure rate for an experimental treatment, Treatment X, and a Standard treatment. She recruits

300 patients and uses a 2:1 randomization, randomizing 200 to receive Treatment X and 100

patients to receive the Standard (because this randomization assigns more to the new

treatment, the researcher can learn more about the safety aspects of Treatment X). After 6

months of treatment, the following cure rates are observed.

Treatment X Standard Treatment

# of patients cured 98 38

n (number of patients) 200 100

Cure Rate 0.49 0.38



=(-0.0076, 0.2276)

Solution:

c.

2 out of 2 points

According to the confidence interval in (b), does it appear that Treatment X has a significantly different cure

rate than the Standard treatment?

Choose: Yes -or- No and provide your rationale.

Your Answer:

No.

Because the 95% confidence interval contains the value 0, we do not have sufficient evidence to

conclude that Treatment X has a siginifcantly diffierent cure rate than the Standard treatment.

Solution:

No, the 95% confidence interval covers 0, so there is not a statistically significant difference in the

cure rates for Treatment X versus the Standard treatment (at the 5% significance level).

a. Note that if H0 is true then p1 = p2=p, where p is the common population cure rate. Give an estimate of this

common population proportion p.

estimate of p = _____________________ .

Question 4

(Chapter 12)

Comparing Treatment X and a Standard Treatment — The Next Analysis:

A researcher wishes to compare the cure rate for an experimental treatment, Treatment X, and

a Standard treatment. She recruits 300 patients and uses a 2:1 randomization, randomizing

200 to receive Treatment X and 100 patients to receive the Standard (because this

randomization assigns more to the new treatment, the researcher can learn more about the

safety aspects of Treatment X). After 6 months of treatment, the following cure rates are

observed.

Treatment X Standard Treatment

# of patients cured 98 38

n (number of patients) 200 100

Cure Rate 0.49 0.38

The researcher also wishes to test the following hypotheses: H0: p1 = p2 versus Ha: p1 > p2

using a 5% level of significance.



1 out of 1 points

Your Answer:

= (n1 1 +n2 2) / (n1 + n2)

= (200*.49 +100*.38) / 300

= .453

Solution:

The estimate of the common population proportion p is = 0.4533.

b.

5 out of 5 points

Conduct the test of the hypotheses in part (a). Provide the test statistic and p-value (including a sketch of

the p-value). Please show your work.

Test Statistic Value = ________________

p-value: ______________________

Your Answer:

z = ( 1- 2) / sqrt( (1- )((1/n1)+(1/n2)))

z = (.49-.38) / sqrt( .453(1-.453)((1/200)+(1/100)))



z = 1.804

Area for the left of z = 1.804 is .9641

p = 1-.9641 = .0359

Solution:

Test Statistic Value = 1.804281

p-value: 0.0359

c.

0.5 out of 1points

Feedback: Those checks are for the CI scenario. In

the HT case, we use phat from part a.

If H0 is true what distribution allows for the calculation of a p-value?

Your Answer:

It is a random and independent sample, and that n1 1, n1(1- 1), n2 2, n2(1- 2) are all greater

than 10.

Model for the z test statistic under Ho is the standard normal (or N(0,1)) distribution was used to

calculate the p-value.

Solution:

The N(0, 1) or standard normal distribution.

d.

0 out of 0.5points

Feedback: This equation is for the binomial

distribution. We have a normal distribution.

If H0 is true what is the expected value of the test statistic?

Your Answer:

E(x)=px=.453(300)=135.9

Solution:

The expected value is the mean, which is 0.

e. Pick the appropriate statistical decision: Reject H0 -or- Fail to Reject H0




Your Answer:

Reject Ho

Solution:

Reject H0

f.

1 out of 1 points

Provide a conclusion in the context of this problem, i.e. regarding the comparison of Treatment X and the

Standard treatment.

Your Answer:

There is sufficient evidence to support that the true proportion of patients who were cured after six

months of Treatment X is greater than the true population proportion of patients cured after six

months of the Standard treatment.

Solution:

We have sufficient evidence to conclude that the population proportion of patients cured after

receiving Treatment X is larger than the population proportion of patients cured after receiving the

standard treatment.

g.


If the decision and conclusion in (e) and (f) are incorrect, it would be a (pick one) Type 1 -or- Type 2 error?

Your Answer:

Type 1

Solution:

Type 1

h.

1 out of 1 points

Describe a consequence of such an error.

Your Answer:

We would conclude that treatment X is more beneficial than the standard treatment when in fact it

is equal or less effective than the standard treatment. As a result of this error, patients may be

encouraged to undergo treatment X, which could actually have the same or lower success rate

than the standard treatment.

Solution:



Solution:

If we reject the null hypothesis when it is actually true, we are incorrectly concluding that

Treatment X results in a higher proportion of cured patients than the Standard Treatment. Since

the reality is that there is no difference between the two treatments, the Type 1 error is likely to

cause investment in a new drug that doesn?t work any better than the standard drug.

a.


Compare the decision made in part (c) of the first question that was based on the confidence interval to the

hypothesis test conclusion made in part (f) of the second question based on the large sample z test.

Are the decisions the same? Pick one:

1. Yes

2. No

Solution: 2. No

b.

1 out of 1 points

Fill in the blanks to provide an answer as to how (or why) this discrepancy could happen.

A confidence interval is like conducting a ___________-sided test whereas the particular hypotheses

being tested in Problem 2 is a ___________ - sided test.

Your Answer:

two, one (right)

Solution:

A confidence interval is like conducting a TWO-sided test whereas the particular hypotheses

being tested in the 2nd problem is a ONE-sided test (to the right).

a. In this situation, which error would be worse?

1. Type 1 error

Question 5

(Chapter 12)

Reflection on the results of the previous two questions: "Comparing Treatment X and a

Standard Treatment" and "Comparing Treatment X and a Standard Treatment - The Next

Analysis".

Question 6

(Chapter 12)

Watches — Consider the following hypotheses:

H0: The watch is waterproof.

Ha: The watch is not waterproof.

Think about what would be a Type 1 and a Type 2 error for these hypotheses.




2. Type 2 error

Solution: 2. Type 2 error

b.

1 out of 1 points

Provide a brief sentence to explain the consequence(s) of such an error.

Your Answer:

A type 2 error would be to assume that the watch is waterproof when it actually is not. This could

cause someone to go into water with the belief that it is waterproof, and break the watch because

it is in fact not waterproof.

Solution:

Here Type 2 error means failing to reject the null hypothesis that the watch is waterproof when the

alternative hypothesis is actually true, i.e. thinking the watch is waterproof when it really is not

waterproof. As a consequence, the watch may be damaged if the buyer uses it as a waterproof

watch after purchasing it.

Return to Dashboard





Documents

STATS 250 HW #4