Practice Exam II

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Problem 1  Please answer the following True or False

To view a plethora of these problems, please go to the link below and check "Probability" and "Confidence Intervals and Z".

http://www.ltcconline.net/greenl/java/Statistics/TrueFalse/statTrueFalseWithChoices2.html

Problem 2  Five percent of all university students are math majors. 

A.  If 80 randomly selected college students are surveyed, what is the exact probability that at least 6 of them will be math majors.

Solution

This is a binomial experiment since the survey question is "Yes" or "No" and the trials are independent.  The cast of characters is

n = 80    p = 0.05    x = 6

We want

P(x > 6) = 1 - P(x <5)

We can put this into the calculator as

1 - binomcdf(80,0.05,5) = 0.210775

Or we can say that there is about a 21% probability that at least 6 of the randomly selected college students will be math majors.

B.  Could you have used the normal distribution to approximate the binomial distribution for part A.

Solution

     No, since np = (80)(0.05) = 4 is less than 5.   

Problem 3

Your tire company's snow and mud tires have an average lifetime of 80,000 miles with a standard deviation of 10,000 miles.  Answer the following assuming the distribution is normal.

A.  If the current guarantee for the tires is 65,000 miles, about what percentage of the tires will wear out before the guarantee expires?

Solution

First find the cast of characters.  We have

m = 80000    s = 10000

We want the probability P(x < 65000). 

Sketch the picture as shown below

 In your TI 83/84 calculator, put in

normalcdf(-99999,65000,80000,10000) = 0.0668

We can say that about 6.7% of the tires will wear out before the guarantee expires.

B.  You want to reconsider the guarantee so that about 98% last past the guarantee period.  What should you set as the guarantee period on your tires?

Solution

This problem asks us to go backwards.  We want to find y such that

P(x > y ) = 0.98. 

The picture shows that we need to subtract from 1.  We find instead

P(x < y) = 0.02

Since we are going backwards, we are finding the inverse normal.  In the calculator, enter

invNorm(0.02,80000,10000) = 59462

We should offer a 59,462 (or about a 60,000) mile guarantee so that about 98% of the tires will last past the guarantee period.  

Problem 4

The Lake Tahoe Visitor's Authority has determined that 65% of the tourists who come to the Lake Tahoe area to go snowboarding are from the Bay area.  The Boarder Motel has all of its 35 rooms booked during this weekend.

A.  Use the normal distribution to estimate the probability that between 20 and 25 of the rooms host bay area visitors?

Solution

We first find the mean and standard deviation.  The formula gives 

m  =  np = (35)(.65)  =  22.75

and

The continuity correction tells us that we want to find

P(19.5 < x < 25.5)

 The picture indicates that we need to find the area between 19.5 and 25.5.  Use the calculator to get

normalcdf(19.5,25.5,22.75,2.82) = 0.7107023161

We can conclude that there is about a 71% chance that between 20 and 25 of the visitors will be from the Bay Area.

B.  Why is your estimate valid?

Solution

We compute np and nq.  We have

        np  =  35(.65)  =  22.75        nq  =  35(.35)  =  12.25

since both of these value are greater than 5, we can conclude that the distribution is approximately normal and our calculations are valid.

Problem 5

 Explain what the difference is between a sampling distribution and the distribution of a sample.

Solution

A sampling distribution is the distribution of all possible samples of a fixed size taken from a population, while the distribution of a sample is the results that occur from only one individual sample that was taken.

Problem 6

It is known that the mean number of houses a Trick-Or-Treater visits is 46 and the standard deviation is 8.  

A.  Assuming that the distribution is approximately normal, what is the probability that your seven year old neighbor will visit fewer than 42 houses on Halloween?

Solution

We have

m = 46    s = 8

We want P(x < 42). 

The picture shows that we need to put into our calcualtor

normalcdf(-99999,42,46,8) = 0.3085375322

We can conclude that there is about a 31% chance that the child will visit fewer than 42 houses.

 B.  25 children were randomly selected and observed.  What is the probability that their mean number of visits is between 48 and 55?

Solution

For this problem, we need to use the sampling distribution with sample size 25.  We have

We want

as the picture below shows

We use the calculator to find

normalcdf(48,55,46,1.6) = 0.1056498297

We can concluded that there is about an 11% chance that mean number of trick-or-treaters will visit between 48 and 55 houses.

Problem 7

Do you favor allowing pilots to carry a gun in the cockpit?  74% of Americans are in favor of allowing pilots to carry a gun in the cockpit. 

A.  80 passengers board a plane heading toward New York.  What is the probability that the greater than 75% of them favor allowing the pilot to carry a gun?  Use the normal approximation to work this problem out.

Solution

First notice that 75% of 80 is

x = (0.75)(80) = 60

Now use the continuity correction.  We want to find the probability

P(x > 60.5)

We calculate the mean and standard deviation as

The picture is shown below

We put in the calculator

normalcdf(60.5,99999,59.2,3.923) = 0.3701791587

We can conclude that there is about a 37% chance that more than 75% of the passengers favor allowing the pilot to carry a gun.

B.  Is the normal approximation valid?  Explain.

Yes, since

np = 60 > 5   and nq = 20 > 5

Problem 8

The manager of Wasabi restaurant tallied the number of customers that he received over a 50 day period.  He found that the mean number per day for this period was 45 with a standard deviation of 8. 

A.  Construct a 95% confidence interval for the true mean. 

Solution

First find the cast of characters:

Now put it into the calculator's TInterval to get:  (42.726,47.274)

B.  Write a sentence that explains your findings.

Solution

We can be 95% confident that the mean number of customers at Wasabi throughout the entire year is between 42.7 and 47.3.

C.  Explain what it means in the context of this study to be 95% confident.

Solution

If many collections of 50 days of customer totals are looked at, different confidence intervals for each would be calculated.  95% of these confidence intervals would contain the true mean customer count for all days.  5% would not contain this mean.

D.  Was it necessary to make any assumptions about the underlying distribution of the population?  Explain.

Solution

No, we do not need to assume normality since n > 30.

Problem 9

The fifteen year survival rate for prostrate cancer is 76%.  A medical researcher has developed a new technique to treat prostrate cancer and has conducted a study on 250 randomly selected men with prostrate cancer who had this new very painful treatment.   Fifteen years after the treatment 210 of these men were still alive.  The researcher wants so find out of the treatment increases the survival rate.

A.  State the Null and Alternative Hypotheses.

Solution

H₀:  p =  0.76

H₁:  p > 0.76

B.  State the repercussions of a Type I error in the context of this study.

Solution

A Type I error means to reject the null hypothesis when the null hypothesis is true.  In this case, that means decide that the survival rate is higher with the treatment when in fact it isn't.  The treatment will end up becoming common practice when it doesn't work.  Countless men will end up going through this painful treatment when in fact it does nothing to help them.

C.  State the repercussions of a Type II error in the context of this study.

Solution

A Type II error means to fail to reject the null hypothesis when in fact it is false.  In this case, this means that there is no evidence to show that the survival rate is higher when in fact it is.  The treatment will be rejected by the FDA and more men will die because this treatment will not be given to them.

D.  Sketch the Rejection Region with a level of significance of 0.05.

Solution

This is a right tailed test.  The z-critical value is 1.645 (using invNormal(0.95) )The Rejection Region is shown below.

E.  Calculate the test statistic and P-value

Solution

We can use our calculator using the 1-PropZTest with

p₀ = 0.77, x = 210, n = 250 and "> p₀"

The test statistic is z = 2.63001826 and the P-value is p = 0.0042690529 and pHat = 0.84.

F.  Use a complete sentence to state your results using a level of significance of 0.05 in the context of the question.

Solution

With a level of significance of 0.05, there is statistically significant evidence to conclude that for all men with prostrate cancer who receive the treatment the survival rate is higher than 76%.

G.  The level of significance of 5% represents a probability.  State what this represents in the context of the study.

Solution

If the treatment does not change the survival rate at all, and if many such studies were done with 250 men with prostrate cancer receiving this treatment, then 5% of these studies would end up falsely concluding that men with prostrate cancer do have a higher survival rate under this new painful treatment.

H.  The P-Value that you obtained represents a probability.  State what this represents in the context of the study.

If the treatment does not change the survival rate at all, and if many such studies were done with 250 men with prostrate cancer receiving this treatment, then only 0.43% of these studies would result in the survival rate for their samples of men to be greater than 84%.

Problem 10

A psychologist is doing research on blindly following orders.  200 volunteers were ordered to push a button that would inflict 50 volts of electricity into a laboratory animal.  35 of them refused to push the button.

A.  Construct a 90% confidence interval for the true proportion of people who will refuse to zap the animal.

Solution

We first find the cast of characters:

n = 200,  x = 35,  cLevel = .90

B.  Write a sentence that explains your findings.

Solution

We can be 90% confident that between 13% and 22% of all people would refuse to push a button to inflict 50 volts of electricity into a laboratory animal if ordered to.

C.  Explain what it means in the context of this study to be 90% confident.

Solution

If many groups of 200 people are tested to see if they would refuse to push the button, different confidence intervals for each would be calculated.  90% of these confidence intervals would contain the true proportion of people who would refuse to push the button when ordered.  10% would not contain this proportion.

D.  Was it necessary to make any assumptions about the underlying distribution of the population?  Explain.

No, since npHat = 35 > 5 and nqHat = 165 > 5.

Problem 11

Nationally, 2% of the population carry a venereal disease.  You are interested in constructing a 95% confidence interval for the mean number of carriers in the Tahoe Basin.  How many people will you need to test if you want a margin of error of  1%?

Solution

First find the cast of characters to find the sample size given a preliminary estimate for p.

p = 0.02,  q = 0.98, E = 0.01, z = .196  (corresponds to a cLevel of .95)

Now plug it into the formula.

So we need to survey 385 people.

Problem 12

Your burger joint just sent out a coupon for fifty cent burgers.  Your research has shown that 20% of coupon bearing customers just purchase a burger resulting a a loss to your restaurant of $0.25, 30% of coupon bearing customers also purchase fries with their burger resulting in a profit of $.50, and the rest opt for the full meal of a burger fries and a drink resulting in a profit of $1.50. 

A.  Write down a probability distribution table for the indicated distribution.

Solution

B.  Find the expected value and standard deviation.

We can do this using the list and stat feature on the calculator.  The calculator gives a mean of 0.85 and a standard deviation of 0.7.

C.  Use  a complete sentence to interpret the expected value in the context of the question.

If many customers visit the burger joint, then the average profit per customer will be about 85 cents.

Problem 13

Suppose the mean number of nights that Americans stay in hotels and motels per year is 7.9 and the standard deviation is 3.1.  A researcher wants to see if this number is different for people who live in South Lake Tahoe.  She surveys 12 randomly selected South Lake Tahoe residents.  Assume the underlying distribution is approximately normal.  The results of the survey are shown below:

   0, 2, 4, 5, 5, 7, 7, 8, 8, 9, 10, 14

Perform the relevant hypothesis test using a level of significance of 0.05 and state your conclusion in the context of the survey.

Solution

First state the null and alternative hypotheses:

Next sketch the picture which is shown below.

Now enter the data into the L1 in the calculator.  Since the standard deviation is known, we can use the z-test.  Be sure to indicate that Data has been inputted instead of stats.  The calculator gives

z = -1.47    p = 0.1412068365

In particular, since p < 0.05 we fail to reject the null hypothesis.  There is insufficient evidence at the 0.05 level of significance to make a conclusion about the mean number of nights that South Lake Tahoe locals stay in hotels and motels per year differs from the national average.

Problem 14

A study was done to determine if the average student get less than the average recommended daily amount of sleep of 8 hours.  The 35 randomly selected students surveyed received an average of 7.6 hours of sleep and their standard deviation was 1.7 hours.  Conduct the relevant hypothesis test using a level of significance of 0.10 and state your conclusion in the context of the survey.

Solution

First state the null and alternative hypotheses.

Next sketch the picture which is shown below

Since the population standard deviation is unknown, we need to us the Student's T-Distribution (T-Test on the calculator).  Now use the calculator with

This gives us a t-statistic of -1.392018772 and a P-Value of 0.086478038.

In particular, since P < 0.10, we can reject the null hypothesis and accept the alternative hypothesis.  There is statistically significant evidence to conclude at a 0.01 level of significance that students on average get less than 8 hours of sleep per day.