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MA2720

Homework 5

Due Tuesday, March 29, 2016

Name_______________________________

Class_______________________________

1. The EPA standard on the amount of suspended solids that can be discharged into rivers and streams is a

maximum of 60 milligrams per liter daily, with a maximum monthly average of 30 milligrams per liter.

Suppose you want to test a randomly selected sample of n water specimens and to estimate the mean daily

rate of pollution produced by a mining operation. If you want a 98% confidence interval estimate of width

2.2 milligrams, how many water specimens would you have to include in your sample? Assume prior

knowledge indicates that pollution readings in water samples taken during a day are approximately

normally distributed with a standard deviation equal to 5 milligrams.

_________________________________(4)

If you can only take 75 water samples and you still need to be 98% confident, what is your new confidence

interval width?

_________________________________(3)

If you can only take 75 samples and you still need to be within plus or minus 1.1 milligrams, what is your

new confidence level?

_________________________________(4)

2. One way corporations raise money for expansions to issue bonds, which are loan agreements to repay the

purchaser a specified amount of money with a fixed rate of interest paid periodically over the life of the bond.

The sale of the bonds is usually handled by an underwriting firm. Does it pay to shop for your bond

underwriter? The reason for the question is that the price of a bond may rise or fall after its issuance.

Therefore, whether a corporation receives the market price for a bond depends on the skill of the underwriter.

Random and independent samples were taken and the mean change in the price ($) of 27 bonds handled over

a 12 month period by one underwriter and in the prices of 23 bonds handled by another are given below.

Underwriter 1 Underwriter 2

Sample Size 27 23

Sample Mean -.0491 -.0307

Sample Variance 0.009800 0.006465

Assuming normality, construct a 99% confidence interval to determine if there is a difference in mean

change in bond prices handled by the two underwriters. (14 points)

Random Variable_____________________

Assumptions:____________________________

_______________________________________

H o _________________________________

H a _________________________________

Reliability Coefficient__________________

(__________________________________)

Conclusion__________________________

____________________________________

Based on your conclusion, what type of error have you subjected yourself to and explain the error in terms

of this particular situation?

____________________________________

_________________________________(4)

3. A pupillometer is a device used to observe changes in an individual’s pupil dilations as he or she is exposed

to different visual stimuli. There is a direct correlation between the amount an individual’s pupil dilates

and his or her interest in the stimuli so marketing organizations sometimes use pupillometers to help them

evaluate potential consumer interest in new products. Suppose 15 consumers were chosen at random and

each was shown two different silverware patterns. The pupillometer readings (mm) for each consumer

are shown in the table below. Assuming normality, test at the 10% significance level to determine if the

mean dilation for pattern 1 exceeds that for pattern 2. (16 points)

Consumer 1 2 3 4 5 6 7 8 9 10 11

Pattern 1 1.00 0.97 1.45 1.21 0.77 1.32 1.81 0.91 0.98 1.46 1.85

Pattern 2 0.80 0.66 1.22 1.00 0.81 1.11 1.30 0.32 0.91 1.10 1.60

Consumer 12 13 14 15

Pattern 1 0.33 1.77 0.85 0.15

Pattern 2 0.21 1.50 0.65 0.05

Random Variable_____________________

Assumptions:____________________________

_______________________________________

H o _________________________________

H a _________________________________

Test Statistic_________________________

Rejection Region______________________

Conclusion__________________________

____________________________________

P-value_____________________________

4. There are many physical and social pressures that lead many urban bus drivers to retire prematurely with

disabilities such as coronary heart disease and stomach disorders. An intervention program was

implemented to improve the work conditions of Stockholm Sweden’s city’s bus drivers and was then

evaluated. A random and independent samples of bus drivers in the intervention program and bus drivers

not in the program were taken and the heart rates (in beats per minute) were determined. Assuming

normality and testing at the 5% significance level, do the data provide sufficient evidence to conclude that

the intervention program reduces mean heart rate of urban bus drivers in Stockholm? (16 points)

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Intervention

68 66 64 58 69 63 68 63 64 71 67 69 60 61

No Intervention

74 62 67 63 77 57 80 77 73 76 54 73 54 60

77 63 76 74 73 54 60 77 83 60 68 64 66 71

Random Variable_____________________

Assumptions:____________________________

_______________________________________

H o _________________________________

H a _________________________________

Test Statistic_________________________

Rejection Region______________________

Conclusion__________________________

____________________________________

P-value_____________________________

If you had tested at the 10% significance level, would your conclusion have changed? Explain.

____________________________________

_________________________________(4)

5. A study examined the reproductive characteristics of the eastern cottonmouth. Random and independent

samples were taken of the number of young per litter for female cottonmouths in Florida and in Virginia.

Preliminary data analyses indicate that you can reasonable presume that litter sizes of cottonmouths in

both states are approximately normally distributed. At the 1% significance level, do the data provide

sufficient evidence to conclude that, on average, the number of young per litter of cottonmouths in Florida

is less than that in Virginia. (10 points)

Group N Mean Std. Dev Std. Error

Florida 24 5.458 1.587 0.324

Virginia 26 7.923 2.883 0.565

If Variances Are t Df Pvalue

Equal -3.70 48 0.000277

Not Equal -3.78 39.5 0.000259

H o _________________________________

H a _________________________________

Assumptions:____________________________

_______________________________________

Test Statistic:________________________

P-Value:____________________________

Conclusion__________________________

____________________________________

____________________________________

代写 MA2720