Chapter 15
Match the following scenarios with the most likely correlation. And why you chose that answer for the following scenario.
Monthly rent and distance from a large city.
Personal Income and personal expenditure.
The amount a student studies and the amount their roommate
A) .8
B) .4
C) -.6
Monthly rent and distance from a large city:
Personal Income and personal expenditure:
The amount a student studies and the amount their roommate studies:
A hospital keeps a record of all kidney transplants done, and finds that 792 transplants have succeeded, and 221 have failed. What is our best guess of the probability of a successful kidney transplant? How can we estimate more accurately?
Probability of success=Number of successful transplants / Total number of transplants
Substituting the given values:
Probability of success=792/ 792+221
Probability of success=792/1013
Probability of success≈0.782
Best guess is .782. To estimate the probability of a successful kidney transplant more accurately, we should gather data from a larger sample size. Considering factors like recipient health, donor organ quality, and medical team expertise can provide more nuanced insights, refining our estimate and making it more accurate.
How would you assign digits to simulate the following scenarios?
(a) Simulating if a baseball player gets a hit in one at-bat if their batting average is .332.
(b) Simulating a student’s response to a question about preferred times to have a class, if, 30% of
students say they prefer morning classes, 50% prefer classes in the afternoon, and 20% prefer
night classes.
(c) Rolling a single fair die.
(a) For simulating a baseball player's at-bat:
(b) For simulating a student's class preference:
(c) For rolling a single fair die:
A November 2018 Pew Research Center survey consisting of a random sample of 4594 adult Americans found that 51% reported using YouTube in order to figure out how to do things they haven't done before.
Find a 95% confidence interval for the proportion of all adult Americans who use YouTube to figure out how to do things they haven't done before. You might find a table of z-critical values useful. Round your answer to three decimal places. Confidence Interval:
Confidence interval=p^±Z × sqrt(p^(1−p^)/n)
Given:
Confidence interval=0.51±1.96× sqrt (0.51(1−0.51)/4594)
Confidence interval=0.51±1.96× sqrt(0.2499/4594)
Confidence interval=0.51±1.96×sqrt(0.0000544)
Confidence interval=0.51±1.96×0.00738
Confidence interval=0.51±0.0145
Therefore, the 95% confidence interval for the proportion of all adult Americans who use YouTube to figure out how to do things they haven't done before is (0.495,0.525)(0.495,0.525).
In 2018, the Pew Research Center published a report in which it was claimed that 44% of adults in the United States prefer watching the news on television as opposed to reading it or listening to it on the radio. A researcher believes that significantly more than 44% of adults prefer watching the news on television. To test his theory, he obtains data from a random sample of 327 adults, and it is observed that 53% of this sample prefers watching the news on television. As a first step in conducting a statistical test of significance, the researcher will write a null hypothesis 𝐻0. Assume that 𝑝 is the population proportion.
In this example, what should be the null hypothesis?
the null hypothesis (H0) is typically a statement of no effect or no difference.
H0: p = 0.44
Describe the relationship between the two correlated variables given as most likely Causation, common response, or confounding. If causation, explain how one variable causes the other. If not, describe a potential lurking variable which can explain the relation.
(a) Number of ice cream cones sold and number of sunburns.
(b) Number of manatees and number of boats registered in a particular body of water
a) Common response - very sunny days causes both ice cream sales and sun burn.
(b) causation - boats hit manatees injuring or killing them
For (a), "Common response" accurately describes the situation where both variables, number of ice cream cones sold and number of sunburns, increase in response to a common factor, which in this case is very sunny days.
For (b), "Causation" correctly identifies the relationship where the number of boats hitting manatees directly causes injury or death to the manatees, thus affecting their population. This relationship implies a direct cause-and-effect link between the two variables.
One of the following statements is sensible, and one is founded in a common myth about probability. Identify which is which:
(a) I believe that after taking a brown M&M out of a bag and eating it, the next M&M I eat is less likely to be brown.
(b) A couple believes that since they have had 4 boys, they are more likely to have a girl because of the law of averages states that the probability of having a boy or girl is roughly equal
Statement (a) is sensible as it reflects the concept of independent events, where each event's outcome does not affect the probability of subsequent events.
Statement (b) is founded in a common myth about probability. It mistakenly assumes that past events influence the probability of future outcomes, which is not true in independent events like childbirth.
An trainee in technical support has the following procedure when answering a call related to problems
with a device: They first ask the the customer to restart their device, which resolves the issue 70% of
the time. If that doesn’t work, they run diagnostic software, which resolves the issue 60% of the time.
If neither works, they will transfer the customer to a specialist. Make a tree diagram to represent this
scenario, and use the digits given to simulate 5 trials and estimate the probability the issue is resolved
during the call.
19254 88013
Tree Diagram:
Restart (70%)
/ \
Diagnostic (30%) Transfer
/ \
Resolve (60%) Specialist
\
No Resolution
Assign digits as follows:
Restarting works: 0-6 Restarting fails:7-9
Software works:0-5 Software fails: 6-9
We get the following results: Yes, No, Yes, Yes, Yes. Estimate probability: 4/5=.8
In 2017, the Pew Research Center published the results of a survey on gun ownership. A random sample of 3930 American adults aged 18 years and older participated in this survey. Of these 3930 individuals, 1269 individuals reported that they owned at least one gun. Of these self‑reported gun owners, 850 said the reason they owned a gun was for protection. Based on this information, determine the sample proportion, p^, of gun owners from this survey who own a gun for protection.
(Use decimal notation. Give your answer to three decimal places.)
p^=
𝑝̂ =count in the sample/size of the sample
=850/1269≈0.670
Suppose that a report by a leading medical organization claims that the healthy human heart beats an average of 72 times per minute. Advances in science have led some researchers to question if the healthy human heart beats an entirely different amount of time, on average, per minute. They obtain pulse rate data from a sample of 85 healthy adults and find the average number of heart beats per minute to be 76, with a standard deviation of 13. Before conducting a statistical test of significance, this outcome needs to be converted to a standard score, or a test statistic. What would that test statistic be? (Use decimal notation. Give your answer to one decimal place.)
test statistic:
First, calculate the standard error. Since the sample size is 𝑛=85 and the sample standard deviation is 𝑠=13, the standard error for the sample of healthy adults is
standard error=𝑠/√𝑛= 13/√85=1.41
Then determine the test statistic. Note that the hypothesized mean is 72 whereas the observed average value is 76. Thus, the test statistic is
test statistic=observation− mean/standard error
=76−72/1.41=2.8
The test statistic would be 2.8.
Eye width and eyelash length are correlated at r=.99 for the scatterplot above. The line of regression for these variables is y = .528x − .158. Use this information to predict the eyelash length of a species whose eye is 1.5cm in width. How confident are you in this prediction? What about a prediction about a species whose eye width is 10cm?
We should feel quite confident about this prediction because of the extremely high
correlation value. We should feel less confident about making predictions about a species whose eye width is 10cm, however, since it falls outside of the range of available data, i.e., we may be extrapolating
You make two piles of 4 cards, numbered 5 through 8. You flip one card from each and add their values, and call the result X.
Find:
(a) the number different outcomes are there for the values of the two cards AND the number of possible values for the sum of the two cards?
(b) P (both cards read “5”)
(c) P (X = 13)
(d) P (X ≠13)
To find the number of different outcomes for the values of the two cards, we consider the combinations of cards from each pile. Since there are 4 cards in each pile, there are 4×4=164×4=16 possible outcomes. Each outcome corresponds to a unique sum of the two cards. For the sum of the two cards, we need to find the range of possible values. The minimum sum occurs when both cards are 5, resulting in 5+5=105+5=10, and the maximum sum occurs when both cards are 8, resulting in 8+8=168+8=16. Therefore, the possible values for the sum of the two cards range from 10 to 16.
(b) Since there is only one card with a value of 5 in each pile, the probability of drawing a 5 from both piles simultaneously is 1/4×1/4=1/16
(c) To find P(X=13), we need to determine the number of outcomes where the sum of the two cards is 13. This occurs when one card is 5 and the other is 8, or vice versa. There are 2 such outcomes. Therefore, P(X=13)=2/8=1/4
d) To find P(X≠13), we need to determine the number of outcomes where the sum of the two cards is not 13. Since there are 2 outcomes that result in 13 (one with 5 and 8, and the other with 8 and 5), there are 8−2=68−2=6 outcomes where the sum is not 13. Therefore, P(X≠13)=6/8=3/4
Calculate the expected value of a game where you roll one fair die and:
Win $4 on an odd roll
Win nothing on a 2 or 4
Lose $6 when rolling a 6.
Would you play this game?
Given:
Now, let's calculate the expected value:
EV=(3/6×4)+(2/6×0)+(16×(−6))
EV=(12/6)+(0/6)+(−6/6)
EV=2−1
EV=1
So, the expected value of the game is $1.
I would recommend playing this game, since the expected value
is positive.
A factory employee believes a machine which manufactures steel rods to be malfunctioning. For a
functioning machine, we expect that only two percent of rods will be outside of the acceptable length, but out of the last 100 rods tested from this machine, 5 were over the acceptable length. Do a hypothesis test to determine if the machine is really malfunctioning.
(a) Identify the population, variable, parameter and sample evidence in the problem.
(b) Determine H0 and Ha.
(c) What is the evidence?
(d) The P -value was 0.0159. Is this result significant at the α = .05 level? Explain what this means.
(a)
b) the null hypothesis (H0) is that the machine is functioning properly, meaning that the proportion of rods outside the acceptable length is equal to or less than 2% (p = 0.02).
H0 : p = .02, Ha : p > .02
c) The evidence is the observed proportion of rods outside of the acceptable length, which is 5 out of 100, or 0.05.
d) Yes - our P -value is smaller than α, so the event is unlikely given the null hypothesis is true, so we believe the alternate hypothesis instead.
A poll which received 1000 results finds that 25% of Americans believe that astrology can make accurate
predictions about ones life. Write a 90% confidence interval for this result. Approximately what are
the odds that the true percentage of Americans who believe in Astrology is greater than the high end
of your interval?
Margin of Error=1.645× sqr(0.25(1−0.25)/1000)
Margin of Error≈1.645× sqr(0.1875/1000)
Margin of Error≈1.645× sqrt(0.0001875)
Margin of Error≈1.645×0.0137
Margin of Error≈0.0226
Now, we can construct the confidence interval:
Confidence Interval=0.25±0.0226
Confidence Interval≈(0.2274,0.2726)
So, the 90% confidence interval for the proportion of Americans who believe in astrology is approximately (22.74%, 27.26%).
The correlation between percentage voting Democrat in 1980 and percentage voting Democrat in 1984 is r=0.704. The correlation between percentage of high school seniors taking the SAT and average SAT Mathematics score in the states is r=-0.86.
Which of these two correlations indicates a stronger straight-line relationship? Explain your answer.
a) The correlation r = -0.86 is stronger because the absolute value of -0.86 is closer to 1 than is the absolute value of 0.704.
b) The correlation r=0.704 is stronger because the absolute value of 0.704 is closer to 0 than the absolute value of -0.86.
c) The correlation r=0.704 is stronger because 0.704 is larger than -0.86.
d) The correlation r= -0.86 is stronger because -0.86 is smaller than 0.704.
a) The correlation r = -0.86 is stronger because the absolute value of -0.86 is closer to 1 than is the absolute value of 0.704.
ou poll a town for their support of a political candidate by taking an SRS of size 100 and record the
proportion who say they will vote for the candidate. You then repeat this poll many times, and find
that the responses are roughly normally distributed with mean .65 and standard deviation .01.
(a) What are the odds of the next poll being between .66 and .68?
(b) What is the probability you conduct this poll and get more than 67.5% of respondents supporting
the candidate?
z=x−μ/σ
a) 68% or .68
z=0.68−0.65/0.01 = 0.025/0.01=2.5
Using a standard normal distribution table or calculator, we find that the probability associated with a z-score of 2.5 is approximately 0.0062, or 0.62%
Suppose that 80% of American Airline flights are on time (this is approximately the percentage of American Airline flights on time in November 2018). You check 10 American Airline flights chosen at random. What is the probability that all 10 are on time?
Give a probability model for checking 10 flights chosen independently of each other. Select the best description of the model.
a) Each flight has a probability of 0.80 of being on time, and 0.20 chance of being delayed. Each flight is not independent of each other.
b) Each flight has a probability of 0.80 of being on time, and 0.20 chance of being delayed. Each flight is independent of each other.
c)Each flight has a probability of 0.08 of being on time, and 0.02 chance of being delayed. Each flight is independent of each other.
d) Each flight has a probability of 0.08 of being on time, and 0.92 chance of being delayed. Each flight is independent of each other.
Since the probability of a flight being on time is 80%, the probability in decimal form is 0.80. The probability of delay (or cancellation) is the complement of 80% or 20%. Therefore, the probability of delay is 0.20. We can check that these two outcomes represent a valid probability distribution if they sum to one.
0.80+0.20=10.80+0.20=1
The problem states that the outcomes are independent, so that the outcome of one flight does not affect the outcome of another. This is reasonable in a randomly selected sample even though, in the real world, cancelled or delayed flights due to weather might occur in bunches. This is because the random sample is not choosing consecutive flights.
c) is correct
A random sample of 197 12th-grade students from across the United States was surveyed and it was observed that these students spent an average of 23.5 hours on the computer per week, with a standard deviation of 8.7 hours. Suppose that you plan to use this data to construct a 99% confident interval. Determine the margin of error.
(Use the table of critical values of the Normal distributions. Usc decimal notation. Give your answer to two decimal places.)
margin of error:
the formula for the margin of error is
margin of error= z * 2/ sqrt(n)
margin of error= 2.58 * 8.7/ sqrt(197) ≈1.60
Thus, the margin of error equals 1.60.
A social psychologist reports, "In our sample, ethnocentrism was significantly higher (P < 0.05) among church attenders than among non-attenders."
Explain to someone who knows no statistics what this means.
a) This means that the study found evidence that church attenders have higher levels of ethnocentrism than non-attenders, which occurs just by chance.
b)This means that there was a 5% difference in ethnocentrism between church attenders and non-attenders and this difference did not occur by chance.
c)This means that the study found important differences in ethnocentrism between church attenders and non-attenders.
d)This means that the study found evidence that church attenders have higher levels of ethnocentrism than non-attenders, which does not occur just by chance.
This means that the study found evidence that church attenders have higher levels of ethnocentrism than non-attenders, which does not occur just by chance.
Fred keeps his savings in his mattress. He began with $1000 from his mother and adds $250 each year. His total savings y after x years are given by the equation y = 1000 + 250x
Draw a graph of this equation. (Choose two values of x, such as O and 10. Compute the corresponding values of y from the equation. Plot these two points on graph paper and draw a straight line joining them.)
After 20 years, how much will Fred have in his mattress? (Give your answer as an exact number.)
a) Fred's savings:
If Fred had added $300 instead of $250 each year to his initial $1000, the equation that describes his savings after x years is
equation:
a)
To calculate how much Fred will have in his mattress, you can use the graph or substitute 𝑥=20 into the equation.
1000+250⋅20=6000
So, after 20 years, Fred will have $6000 in his mattress.
b)
If Fred had added $300 instead of $250 each year to his initial $1000, the new equation would have the new slope, 300,300, and the same intercept as the old equation. Thus, the equation that describes Fred's savings after 𝑥 years if he had added $300 instead of $250 each year to his initial $1000 is
equation: 𝑦=1000+300𝑥
Finding the number of standard deviations: You correctly calculated that 40.8% is two standard deviations below the mean of 44%.
Using the 68-95-99.7 rule: By applying this rule, you determined that 95% of the population falls within two standard deviations of the mean. Therefore, 5% of the population is outside this range, with half below 40.8% and half above 47.2%.
Calculating the probability: Adding the 95% within two standard deviations to the 2.5% above 47.2%, you correctly concluded that there is a 97.5% chance of being above 40.8%.
Winning numbers for the "Straight" DC-4 lottery game are reported on television and in local newspapers. You pay $1.00 and pick a four-digit number. The state chooses a four-digit number at random and pays you $5000 if your number is chosen.
What are the expected winnings from a $1.00 Straight DC-4 wager?
Please round your final answer to the nearest cent.
The probability of winning the Straight DC-4 lottery is 1 out of 10,000 (since there are 10,000 possible four-digit numbers).
So, the expected winnings can be calculated as follows: Expected Winnings=Probability of Winning×Amount Won−Cost of Wager
Expected Winnings=1/10,000×$5000−$1.00
Expected Winnings=$0.50−$1.00
Expected Winnings=−$0.50
So, the expected winnings from a $1.00 Straight DC-4 wager is approximately -$0.50. This means, on average, you would lose 50 cents per wager.
In 2016, the Pew Research Center reported the results of a survey about public library usage. A nationally representative sample of 1601 people aged 16 and older who were living in the United States completed the survey. One survey question asked whether the survey respondent had visited, in person, a library or a bookmobile in the last year. A total of 768 of those surveyed answered "yes" to this question. Although the samples in national polls are not simple random samples, they are similar enough that our method gives approximately correct confidence intervals. Give a 90% confidence interval for the proportion of all American adults who answered yes on the survey question
(Use the table of critical values of the Normal distributions Give your answer as an interval in the form (lower limit, upper limit). Use decimal notation. Give the limits to two decimal places)
99 confidence interval:
Given:
Now, let's calculate the confidence interval:
Confidence interval=0.480±1.645× sqrt(0.480(1−0.480)/1601)
Confidence interval=0.480±1.645× sqrt(0.2496/1601)
Confidence interval=0.480±1.645× sqrt(0.0001558)
Confidence interval=0.480±1.645×0.012487
Confidence interval=0.480±0.02052
Therefore, the 90% confidence interval for the proportion of all American adults who answered "yes" to the survey question about visiting a library or bookmobile in the last year is approximately (0.459,0.501)(0.459,0.501).
The National Collegiate Athletic Association (NCAA) requires colleges to report the graduation rates of their athletes. At one large university, 82% of all students who entered in 2012 graduated within six years. One hundred forty-nine of the 190 students who entered with athletic scholarships graduated. Consider these 190 as a sample of all athletes who will be admitted under present policies. Is there evidence that the percentage of athletes who graduate is less than 82%?
What are the null and alternative hypotheses Ho and Ha
a) Ho:p=0.82
Ha:P <0.82
b) Ho: p=0.82
Ha:p > 0.82
c) Ho:p=0.82
Ha:p <0.82
d) Ho:p=0.784
Ha:p <0.784
e) Ho: P = 0.784
Ha:P <0.784
f) Ho:p = 0.82
Ha: p > 0.82
Null Hypothesis (H0):
Alternative Hypothesis (Ha):
a) Ho:p=0.82
Ha:P <0.82