HW #5 Solutions

Assignment Stats 250 W11 HW 5

Graded Point Total: 25.5 out of 30

a.

0 out of 0.5points

Feedback:

This is the expected value of xbar, not xbar. This is

one of those important differences between a

parameter and a statistic.

See Textbook for Exercise: 9.a.

Your Answer:

=7.05 hours

Solution:

The mean would be the population mean µ = 7.05 hours

b.

0.5 out of 1points

Feedback: Same as above.

See Textbook for Exercise: 9.b.

Your Answer:

s=σ/sqrt(n)

s=1.75/sqrt(190)=.1270

Solution:

s.d.( ) = σ/sqrt(n) = 1.75/sqrt(190) = 0.127 hours

c.

1 out of 1 points

See Textbook for Exercise: 9.c.

Your Answer:

between 6.923 and 7.177 hours

Solution:

6.923 and 7.177, calculated by going out one standard deviation each way from the mean,

namely, 7.05 ± (1 x 0.127).

a. Consider the procedure of selecting a random sample of 64 salaries and computing the sample mean

salary, . Suppose you repeated that procedure many, many times. Describe and draw the distribution of

the resulting values. Also provide the values along the axis that represent 1, 2, and 3 standard

deviations from the mean.

Required Assignment Stats 250 W11 HW 5

Question 1

(9.57)

See Textbook for Exercise: 9.57.

Question 2

(Chapter 9)

Let X = the salary (in $1000s) for a randomly selected full-time worker in a certain city. Suppose

the expected (or mean) salary is $62.0 and the standard deviation is $12.0.

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

lecturebook.com/homeworks/show/41 1/9

3 out of 3 points

Your Answer:

Because the sample size is 64,which is greater than the required 25 for the Central Limit Theorem, is approximately N(62,1.5) ba

Limit Theorem.

Solution:

b.

2 out of 2 points

If you were to obtain a random sample of 64 salaries, what is the probability that the sample mean salary

will exceed $65?

Your Answer:

z score = (65-62)/1.5 = 2

P(z>2) = 1 - P(z<2) = 1 - .9772 = 0.0228

The probability that the sample mean salary will exceed $65 is 0.0228.

Solution:

P( > $65) = P(Z > ($65 - $62)/$1.50) = P(Z > 2) = 1-P(Z ≤ 2) = 1-0.9772 = 0.0228

c.

0.5 out of 1points

Feedback: We don't have a sample yet in this type of

question. We need to know about the population.

Could you apply the same technique with a random sample of 9 salaries? Explain using 1 brief sentence.

Your Answer:

No, because with a sample size of 9 is not large enough of a sample to apply the Central Limit

Theorem, and because the prompt does not say it is a normal sample, we would have to check

for normality first.

Solution:

N W t i th d l di t ib ti f l i f thi l ti l h th



No. We are not given the model or distribution for salaries for this population, we only have the

mean and standard deviation. Thus we cannot apply the same technique because we do not

have a large enough sample size (e.g. n > 25 or 30) to rely on the Central Limit Theorem, as we

did to answer parts (a) and (b).

0 out of 1 points

a.

1 out of 1 points

Provide a 95% confidence interval for µ.

Your Answer:

=18.25

s=3.55

s.e.( )=3.55/sqrt(18)=.8367

df = n-1 = 18-1 = 17

95% confidence interval:

+/-t*s.e.( ) = 18.25 +/- 2.11*.8367

=(16.4846,20.0154)

Solution:

± t*s.e.( ) --> $18.25 ± 2.11(0.8367) -> $18.25 ± $1.77

where the standard error of the sample mean is s.e.( ) = 3.55/sqrt(18) = 0.8367

So the 95% Confidence Interval is given by ($16.48 , $20.02).

b.

0 out of 1 pointsFeedback: You've interpreted the interval, not the

level.

Interpret the 95% confidence level.

Your Answer:

We are 95% confident that the true mean of all UM students carry an amount of cash inbetween

the interval (16.4846,20.0154)

Solution:

If we repeatedly took a random sample of 18 UM students from this same population and for each

sample compute the 95% CI for the population mean "cash-in-pocket", we would expect 95% of

those resulting confidence intervals to contain the true population mean "cash-in-pocket".

Question 3

(Chapter 9)

Is the following statement true or false?

If the model for scores on a placement test for incoming Freshmen students is strongly skewed

left, and a large number of such students (say n = 200) are selected randomly, then the model for

Freshmen placement test scores will become approximately normal.

1. TRUE

2. FALSE

Solution: 2. FALSE

Question 4

(Chapter 11)

Suppose you obtained a random sample of 18 UM students and measured the amount of cash

they were carrying with them, called "cash-in-pocket". The resulting mean is $18.25 and the

standard deviation is $3.55. Let µ = the population mean amount of "cash-in-pocket" for all UM

students.



c.

1 out of 1 points

If you were to make a 99% confidence interval instead,

it would ______________ (pick one):

1. be wider

2. be narrower

3. remain unchanged

Solution: 1. be wider

3 out of 3 points

a.

1 out of 1 points

If the null hypothesis is true, what is the distribution of the test statistic?

Question 5

(Chapter 11)

Think about the formula for making a confidence interval to estimate the population mean. Name

the three factors that affect the width of a confidence interval and for each factor, indicate how an

increase in the numerical value of the factor affects the interval width. Use the following template

three times for your answer.

"An increase in the ____________ would make the interval ___________ (select wider or

narrower)."

Your Answer:

An increase in the confidence level would make the interval wider

An increase in the sample size would make the interval narrower

An increase in the sample standard deviation would make the interval wider

Solution:

1. An increase in the confidence level would make the interval wider.

2. An increase in the sample size would make the interval narrower.

3. An increase in the sample standard deviation would make the interval wider.

Question 6

(Chapter 13)

Does Drug X Significantly Lower Blood Pressure? You are on a team of scientists that is trying

to obtain approval for a new drug (call it Drug X) that should lower diastolic blood pressure for

hypertensive patients. Hypertensive patients have diastolic blood pressures that are much

greater than 90 mmHg.

The team has designed a study and will test H0: µ = 90 versus Ha: µ < 90 where µ is defined to

be the average diastolic blood pressure for the population of hypertensive patients treated with

Drug X. A 5% significance level was specified before the study was initiated. You would like to

reject H0 which would imply the elevated blood pressures are brought into "control" (defined as

being less than 90).

The random sample of 30 hypertensive patients was treated and some results from SPSS are

provided below.



Your Answer:

df = N-1 = 30-1 = 29

The distribution would be t(29)

Solution:

Answer = t(29)

b.

0.5 out of 0.5points

If the null hypothesis is true, what is the expected value of the test statistic?

Your Answer:

expected t value = 0

Solution:

Answer = 0

c.

1 out of 1 points

Fill in the blank:

The sample mean of 87.7 mmHg was standard errors below the hypothesized mean of 90 mmHg.

Your Answer:

1.615

Solution:

The sample mean of 87.7 mmHg was <u> 1.615 </u> standard errors <u>below</u> the

hypothesized mean of 90 mmHg.

d.

1 out of 1 points

The Vice President has asked you to summarize the findings of this study with a brief well-written

paragraph as to your conclusion which will facilitate appropriate further development of Drug X. The memo

is confidential and expected to be on the Vice President?s desk by tomorrow morning. One item that is

needed to include in this summary is the p-value. What is that value?

The p-value is:

Your Answer:

p-value is .118/2 = .059

Solution:

The one-sided to the right p-value will be 0.118/2 = 0.059, which is just greater than 0.05.

e.

1 out of 1 points

Which of the following statements is most appropriate to include in your paragraph summary? Pick one:

1. There was insufficient evidence to show the average diastolic blood pressure was

sufficiently lowered below an average of 90 mmHg. Hence, further development of Drug X

should be carefully evaluated and possibly curtailed.

2. Although there was insufficient evidence to show the average diastolic blood

pressure was sufficiently lowered at the 5% level, the results do approach significance.

Hence, further development of Drug X could be considered, possibly fine-tuning the dose or

considering another study with more patients.

3. Drug X significantly lowered average diastolic blood pressure below 90 mmHg.

Hence, we should proceed with an application for approval to the FDA as soon as possible.



Solution: 2. Although there was insufficient evidence to show the average diastolic blood

pressure was sufficiently lowered at the 5% level, the results do approach significance. Hence,

further development of Drug X could be considered, possibly fine-tuning the dose or considering

another study with more patients.

a.

1 out of 1 points

Compute the value of the test statistic

Your Answer:

t= -µo/(s/sqrt(n)) = (162-160) / (6.7/sqrt(41)) = 1.911

Solution:

The test statistic formula is .

The value for the test statistic is .

Test Statistic = t = 1.91

b.

2 out of 2 points

Please upload a sketch showing the appropriate area that corresponds to the p-value, and then report the

p-value.

Your Answer:

df = 40, t= 1.911, so it is between 1.80 and 2.00 on the chart.

Therefore, the p-value is between 0.026 and 0.040.

Solution:

Question 7

(Chapter 13)

The tensile strength for a standard hinge is 160 lbs. A new material is being considered that

claims to have an average breaking strength that exceeds 160 lbs. Let µ = the population mean

breaking strength for all hinges made from the new material. You are asked to test H0: µ = 160

versus Ha: µ > 160 with a significance level α = 0.05. A random sample of 41 hinges made with

the new material is obtained and the resulting mean breaking strength was 162 lbs and the

standard deviation was 6.7 lbs.



According to Table A.3 the p-value is between 0.026 and 0.040

c.

1 out of 1 points

Which is the appropriate statistical decision & conclusion.

i. Reject H0; there is insufficient evidence to demonstrate the new material was acceptable.

ii. Reject H0; the new material does appear to have a mean breaking strength significantly greater than

160 lbs.

iii. Fail to reject H0; there is insufficient evidence to demonstrate the new material was acceptable.

iv. Fail to reject H0; the new material does appear to have a mean breaking strength significantly greater

than 160 lbs.

Your Answer:

ii. Reject H0; the new material does appear to have a mean breaking strength significantly greater

than 160 lbs.

Solution:

Reject H0; the new material does appear to have a mean breaking strength significantly greater

than 160 lbs.

d.

0.5 out of 1points

One assumption for this test to be valid is that we have a random sample from the population of interest.

There is another assumption which can be assessed by examining a histogram and a QQ plot. Clearly

state the other assumption.

Your Answer:

The other assumption that the sample is approximately normally distributed.

Solution:

The remaining assumption is that the population of breaking strengths has a normal distribution.

e.

1 out of 1 points

Suppose the histogram and QQ plot indicate a skewed to the left distribution. Does this imply the test

results are no longer valid? Explain.

Your Answer:

No, because the sample size is greater than 25, so the Central Limitation Theory states that the

distribution is approximately normal.

Solution:

No. Since our sample is large enough (n = 41 which is greater than 25), the Central Limit

Theorem tell us that the sampling distribution of the sample mean is approximately normal even

if the parent population is skewed.



a.

0.5 out of 1points

Feedback: We don't assume we know mu.

Give an interpretation the standard error in context with this situation.

Your Answer:

The standard error estimates, approximately, the average distance of the possible sample mean

temperatures is .0161 degrees fahrenheit from the true population mean temperature of 98.6

degrees fahrenheit.

Solution:

We would estimate the average distance between the possible sample mean values and the

population mean to be about 0.161 degrees. (Note: the possible sample mean values would

come from taking many random samples of size 18 from the same population.)

b.

1 out of 1 points

Based on the provided data, the observed test statistic was t = -2.38 and corresponding p-value was

0.015. Interpret this p-value in context with this situation.

Your Answer:

Given the null hypothesis of µ=98.6 degrees fahrenheit is true and this procedure was repeated

many times, the p value of .015 means the probability of seeing a test statistic as extreme or more

extreme than the observed t=-2.38 is 0.015. This means the sample mean of 98.217 is

somewhat unusual.

Solution:

If the population mean body temperature is 98.6 degrees, and if repeated samples of 18

temperatures are obtained, then we would see a t test statistic of -2.38 or even smaller in only

1.5% of the repetitions.

Or

If the population mean body temperature is 98.6 degrees, the probabiltiy of seeing a t test statistic

of -2.38 or even smaller is only 0.015.

1 out of 1 points

Question 8

(Chapter 13)

A Few Review Questions about a Population Mean μ. Consider the Normal Human Body

Temperature Example on pages 553-558 (Section 13.2). Defining μ = the population mean body

temperature in all humans, the hypotheses to be tested are H0: μ = 98.6° versus Ha: μ < 98.6°. A

random sample of 18 normal body temperatures resulted in a sample mean of 98.217° and a

standard error of the sample mean of 0.161°.

Question 9

(Chapter 13)

Is the following statement True or False? Hypotheses and conclusions from hypothesis testing

apply only to the samples on which they are based.

1. True

2. False

Solution: 2. False

Question 10

(Chapter 13)

Is the following statement True or False?

The p-value is a probability that the null hypothesis is true.

1. True

2. False



1 out of 1 points

Solution: 2. False

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HW #5 Solutions