Intro to stats final

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Renee scores an average of 153 points in a game of bowling, and her points in a game of bowling are normally distributed. Suppose Renee scores 175 points in a game, and this value has a z-score of 2. What is the standard deviation?

11 (back substitute) 2=175-153/symbol for standard deviation (sigma) 2=22/sigma 2sigma=22 sigma=11, so the standard deviation is 11

A group of students were surveyed about the number of siblings they have. The relative frequencies of their responses are shown in the table below. Complete the cumulative relative frequency table Number of siblings relative frequency 0 0.18 1 0.33 2 0.16 3 0.14 4+ 0.19

cumulative relative frequency 0.18 0.51 0.67 0.81 1.0 (add the number with the one before it)

Juan wants to estimate the percentage of people who will vote for issue A on the next election. He surveys 150 individuals and finds that 81 will vote for issue A. Find the margin of error for the confidence interval for the population proportion with a 90% confidence level.

margin of error= .067 p'= 81/150=.54 Substitute the given values p′=0.54, n=150, and zα/2=1.645 to get .067

Greg wants to know the mean of his test scores, which are listed below. 78,82,95,88,82 Find the mean test score.

mean=85 points add them all up and divide by the amount of values (5)

Find the median of the following list of points deducted from a random sample of chemistry final exams. 26,13,39,22,42,39,18,21,32,14

median= 24 points line them up and take the middle value. in this case there were 2 middle values, those get added up and divided by 2

A casino features a game in which a weighted coin is tossed several times. The table shows the probability of each payout amount. To the nearest dollar, what is expected payout of the game? payout amount probability $150 0.161 $2800 0.03 $105000 0.0006

$171 To find the expected value, multiply each payout amount by its probability and round to the nearest dollar: $150(0.161)+$2800(0.03)+$105000(0.0006)≈$171.

In a survey, a random sample of families were asked whether they were planning to take a vacation within the next year. The survey resulted in a sample proportion of p^=0.6, with a sampling standard deviation of σp^=0.09, who stated they were taking a vacation. Using the Empirical Rule, write a 95% confidence interval for the true proportion of families taking a vacation.

(0.42,0.78) ;(p−z⋅σ, p+z⋅σ)=(0.6−0.18,0.6+0.18)(0.42,0.78)

A statistics professor recently graded final exams for students in her introductory statistics course. In a review of her grading, she found the mean score out of 100 points was a x¯=77, with a margin of error of 10.

(67, 87) (point estimate-margin of error, point estimate+margin of error)= (77-10, 77+10)= (67,87)

A teacher claims that the proportion of students expected to pass an exam is greater than 80%. To test this claim, the teacher administers the test to 200 random students and determines that 151 students pass the exam. The following is the setup for this hypothesis test: {Ho:p=0.80 {Ha:p>0.80 Find the test statistic for this hypothesis test for a proportion.

-1.59 proportion of successes=p^=proportion; 151/200=.755 plug numbers into formula: .755-.80/ the square root of .80(1-.80)/200 =-1.59

Suppose the number of inches of rainfall each year in a city is normally distributed. For a random sample of years, the confidence interval (3.9,7.7) is generated. Find the margin of error

1.9; subtract the lower from the upper and divide by 2= 7.7−3.9/2=1.9

According to a survey, the average American person watches TV for 3 hours per week. To test if the amount of TV in New York City is less than the national average, a researcher decides to do a hypothesis test, at a 1% significance level. She surveys 19 New Yorkers randomly and asks them about their amount of TV each week, on average. From the data, the sample mean time is 2.5 hours per week, and the sample standard deviation (s) is 0.9 hours. Ho: μ≥3; Ha: μ<3. α=0.01 (significance level) What is the test statistic (t-value) of this one-mean hypothesis test (with σ unknown)?

-2.42 sample mean-compared= numerator σ/ square root of n=denominator 2.5-3/0.9/ square root of 19= -0.5/0.206= -2.42

Two friends are opening a coffee shop. As they write their business plan, they research the amount of debt similar businesses can have in the first two years of opening. It is known that 72% of coffee shops have a debt of over $50,000 within the first two years of opening. If a random sample of 36 coffee shops is obtained, what is the probability that more than half of them had debt of over $50,000 within the first two years of opening?

.998 plus and minus 0.001 We are given a population proportion of p=0.72. For samples of size n=36, we want to find P(p^>0.50), the probability that the sample proportion p^ will be more than 50%=0.50. Substitute μp^=0.72 and n=36 to get 0.0748 then use the zscore formula .50-.72/.0748= -2.94 find this value on zscore chart to get .002 1-.002 is .998

From a recent company survey, it is known that the proportion of employees older than 55 and considering retirement is 8%. For a random sample of size 110, what is standard deviation for the sampling distribution of the sample proportions,

0.026 plug in that the population proportion p=8%=0.08 and a sample size of n=110 to get .026

Determine the area under the standard normal curve that lies to the right of the z-score 0.05 and to the left of the z-score 0.25.

0.0788 From the given table, we find that P(Z<0.25)=0.5987 and P(Z<0.05)=0.5199. We can take the area to the left of z=0.05 and subtract it from the area to the left of z=0.25, to find what is left in between. P(0.05<Z<0.25)=P(Z<0.25)−P(Z<0.05)=0.5987−0.5199=0.0788

Suppose speeds of vehicles traveling on a highway have an unknown distribution with mean 63 and standard deviation 4 miles per hour. A sample of size n=44 is randomly taken from the population and the mean is taken. Using the Central Limit Theorem for Means, what is the standard deviation for the sample mean distribution?

0.60 4/the square root of 44 =0.60

Identify the parameter p in the following binomial distribution scenario. The probability of buying a movie ticket with a popcorn coupon is 0.711 and without a popcorn coupon is 0.289. If you buy 21 movie tickets, we want to know the probability that more than 14 of the tickets have popcorn coupons. (Consider tickets with popcorn coupons as successes in the binomial distribution.)

0.711 The parameters p and n represent the probability of success on any given trial and the total number of trials, respectively. In this case, success is a movie ticket with a popcorn coupon, so p=0.711.

Suppose a bowler claims that her bowling score is not equal to 150 points, on average. Several of her teammates do not believe her, so the bowler decides to do a hypothesis test, at a 5% significance level, to persuade them. She bowls 22 games. The mean score of the sample games is 157 points. The bowler knows from experience that the standard deviation for her bowling score is 18 points. Ho: μ=150; Ha: μ≠150α=0.05 (significance level) What is the test statistic (z-score) of this one-mean hypothesis test?

1.82 sample mean-population mean= numerator σ/square root of n= denominator 7/18/√22≈1.82

The acts in a talent competition consist of 4 instrumentalists, 10 singers, and 6 dancers. If the acts are ordered randomly, what is the probability that a dancer performs first?

3/10 There are 6 dancers and a total of 4+10+6=20 acts. So, the probability of a dancer being selected to perform first is 6/20=3/10.

The number of square feet per house are normally distributed with a population standard deviation of 137 square feet and an unknown population mean. A random sample of 19 houses is taken and results in a sample mean of 1350 square feet. Find the margin of error for a 80% confidence interval for the population mean.

40.29; since the confidence interval here is 80, you take 1-.80 which is .20 and divide by 2 to get 0.10, the z value associated with this is 1.282. plug that σ=137, n=19, and the previously calculated z value into the formula to get 1.282x137/square room of 19 to get 40.29

Doris is investigating if height has any effect on red blood cell count. What is the explanatory variable?

height

The total snowfall per year in Laytonville is normally distributed with mean 99 inches and standard deviation 14 inches. Based on the Empirical Rule, what is the probability that in a randomly selected year, the snowfall was less than 127 inches? Enter your answer as a percent rounded to 2 decimal places if necessary.

97.5% Notice that 127 inches is two standard deviations greater than the mean. Based on the Empirical Rule, 95% of the yearly snowfalls are within two standard deviations of the mean. Since the normal distribution is symmetric, this implies that 2.5% of the yearly snowfalls are greater than two standard deviations above the mean. Alternatively, 97.5% of the yearly snowfalls are less than two standard deviations above the mean.

A company has developed a wristband for monitoring blood sugar levels without requiring direct blood samples. It is interested in demonstrating the accuracy of the device for governmental approval and has decided to test the claim "The glucose level reported by the wristband is within 10% of a standard blood test result." Which of the following data collection processes would be appropriate? Select only one answer choice.

Choose a random sample of people and random times throughout the day and measure their blood sugar using both the wristband and a standard blood test.

Event A: Rolling an even number on a fair die. Event B: Rolling a 6 on a fair die. Event C: Rolling an odd number on a fair die. Given the three events, which of the following statements is true? Select all that apply.

Event B and Event C are mutually exclusive. Event A and Event C are mutually exclusive. Remember that in general, A and B are mutually exclusive events if they cannot occur at the same time. This means that A and B do not share any outcomes and P(A AND B)=0. In this question, note that A={2,4,6} and B={6} have an outcome in common, 6. Therefore, A and B are not mutually exclusive. Similarly, C={1,3,5}, which has no outcomes in common with A. Therefore, A and C are mutually exclusive.

Is the statement below true or false? Continuous is the type of quantitative data that is the result of measuring.

True

A dental assistant is interested in the proportion of patients that need a root canal. Let the proportion of patients that need a root canal be p. If the dental assistant wanted to know if the proportion of patients that need a root canal is more than 20%, what are the null and alternative hypotheses?

Ho: p=0.20; Ha: p>0.20

Which of the following scenarios demonstrates sampling bias?

Jean wants to estimate the mean amount of money spent on clothes per week by mall shoppers. She collects data from every 10th person entering a clothing store at the mall.

Given the following frequency table of values, is the mean or the median likely to be a better measure of the center of the data set? value frequency 23 4 24 3 25 4 26 4 27 2 28 4

Mean bc most of the values are close together

An English professor asks her students who their favorite character is in the novel they are reading. What is the level of measurement of the data?

Nominal

A basketball player has a 0.654 probability of making a free throw. If the player shoots 13 free throws, what is the probability that she makes less than 6 of them? Insert the correct symbol to represent this probability.

P(X≤5)

The general election is today, and volunteers are organizing the ballots. Depending on the part of the city in which a citizen lives, he or she receives a different ballot. The types of ballots consist of RED ballots numbered 1,2,3,4 and BLUE ballots numbered 1,2. Let R be the event of drawing a red ballot, B the event of drawing a blue ballot, E the event of drawing an even numbered ballot, and O the event of drawing an odd ballot. Drawing the Blue 2 is one of the outcomes in which of the following events? Select all correct answers.

R′ B OR O Because the ballot is blue and the number is even, the ballot is an outcome of B and E. Therefore, it is also an outcome of B OR O and R′.

In a study about vitamins, researchers found that most adults take a daily multivitamin. Specifically, the study found that 7 out of 12 adults (or 58% of adults) take a multivitamin. A medical adviser is interested in studying this topic further. He would like to test the claim that the proportion of adults who take a daily multivitamin is more than 0.58 (the proportion from the previous study). He sets up the hypotheses as: Null hypothesis: p=0.58 Alternative hypothesis: p>0.58

Since 13 out of 20 samples have more than 7 adults who take a daily multivitamin, there is evidence to suggest that the proportion of adults who take a multivitamin is more than 58%.

Gail averages 153 points per bowling game with a standard deviation of 14.5 points. Suppose Gail's points per bowling game are normally distributed. Let X= the number of points per bowling game. Then X∼N(153,14.5).

Suppose Gail scores 108 points in the game on Thursday. The z-score when x = 108 is −3.103. The mean is 153. This z-score tells you that x = 108 is 3.103 standard deviations to the left of the mean. so 108-153/14.5= -45/14.5= -3.103

Given the frequency table below, what is the relative frequency of the data value 2? value frequency 2 6 3 4 4 5 5 7 6 3

The answer is 0.24. Add up all the frequencies (25) then take the frequency associated with the given data value (2) and divide 6/25=0.24

Given the plot of normal distributions A and B below, which of the following statements is true? Select all correct answers.

The means of A and B are equal. A has the larger standard deviation. Because A and B are centered at the same point, their means are equal. Because A is shorter and more spread out than B, we find that A has the larger standard deviation.

Find the mode of the following list of points scored by teams participating in a volleyball tournament. 10,10,7,18,11,14,7,11,14,11,18

The mode is 11. this number occurred the most amount of times in the data set

The following data was calculated during a study on weight loss. Interpret the p− value for this hypothesis test.: A doctor would like to test the claim that the average weight loss after participating in a diet program is less than 10 pounds. The test statistic is calculated as z0=−2.50 The p-value is 0.0062.

The probability of observing a value of z0=−2.50 or less if the null hypothesis is true is 0.62%.

Last year, the proportion of tax filers that received a refund was 25%. An accountant believes this upcoming tax year will have a smaller proportion of tax refunds than the proportion from last year. Interested in studying this further, the accountant samples a few of their clients and determines the proportion that receive a refund for this upcoming tax year is 19%. As the accountant sets up a hypothesis test to determine if their belief about this upcoming tax year is correct, what is the accountant's claim?

The proportion of tax filers that receive a refund is less than 25%.

A poll was conducted during the final game of the basketball season to determine whether fans wanted to see the defending champions win the game or the challenging team win the game. From the poll, 216 of the 374 residents sampled from urban areas want the defending champions to win the game. In more rural areas, 304 of the 466 residents polled want the defending champions to win the game. Assuming location has nothing to do with team preference, the probability that the data gathered was the result of chance is calculated to be 0.03. What is the correct interpretation of this calculation?

The results are statistically significant at the 0.05 level of significance in showing that the proportion of people in rural areas who want the defending champions to win the game is different than the proportion of people in urban areas. The probability value calculated is 0.03. This is less than 0.05, so we conclude that the data are statistically significant in showing that the proportion of people in rural areas who want the defending champions to win the game is different than the proportion of people in urban areas.

Karen collected data from a random sample of 1200 homeowners in her state asking whether or not they use electric heat. Based on the results, she reports that 50% of the homeowners in the nation use electric heat. Why is this statistic misleading?

The sample is biased.

Is a survey that asks the opinions of students about a proposed tour program to a hill station an observational study or experimental study? If it is an experimental study, what is the controlled factor?

The study is an observational study.

Identify the Type II error if the null hypothesis, H0, is: The capacity of Anna's car gas tank is 10 gallons. And, the alternative hypothesis, Ha, is: Anna believes the capacity of her car's gas tank is not 10 gallons.

There is insufficient evidence to conclude that the capacity of her car's gas tank is not 10 gallons when, in fact, it is not 10 gallons.

An SEO account manager is concerned website developers are using too many keywords per web page. The SEO account manager would like to carry out a hypothesis test and test the claim that a web page has, on average, more than 10 keywords. Why is this hypothesis test right-tailed?

This is a right-tailed test because a direction is specified. The population parameter is greater than the specified value.

5% of adults participate in at least 30 minutes of exercise each day. How likely is it that a randomly chosen adult will exercise 30 minutes each day?

Unlikely, the probability is close to 0. It is not closer to 1 than it is to 0, so it is not appropriate to call it "likely" or "somewhat likely." It is very closer to 0 than it is to 0.5, so it is not appropriate to call it "somewhat unlikely". It is false to call it "equally likely" since the probability isn't 0.5. We settle on "unlikely," since it is much closer to 0 than to 1.

The speeds of vehicles traveling on a highway are normally distributed with an unknown population mean and standard deviation. A random sample of 11 vehicles is taken and results in a sample mean of 65 miles per hour and sample standard deviation of 9 miles per hour. Find a 90% confidence interval estimate for the population mean using the Student's t-distribution.

We are 90% confident that the true population mean lies between 60.08 and 69.92 sample-1 and 1-% divided by 2 to get t value. 11-1=10 and 1-.90=.10 and .10/2=.05 so the tvalue is 1.812. plug t=1.812, s=9 and n=11 into formula to get 4.92. confidence interval can be written as (65-4.92, 65+4.92) or (60.08, 69.92)

Andrea conducted a survey in which she collected data on the percentage of people who like horror films and the percentage of people who like romantic dramas. Which of the following could sufficiently display the data if only the two given categories are to be included?

bar graph

The following frequency table summarizes a set of data. What is the five-number summary? value frequency 2 3 3 2 5 1 6 3 7 1 8 2 11 3

min Q1 median Q3 max 2 3 6 8 11 automatically the min is 2 and the max is 11. write out list ex. 2,2,2,3,3,5,6,6,6,7,8,8,11,11,11 6 has 7 values above and below it, so that's the median. with in the first half 3 has 3 values above and below, so thats Q1. in the second half 8 has 3 values above and below it so that's Q3.

In a large population, about 10% of people do not like the taste of cilantro, an herb used in cooking. A researcher takes a random sample of 15 people and surveys whether they like cilantro. Use the binomial distribution to compute the probability that exactly 6 of the people in the sample do not like cilantro. Identify the following information required to find the probability of people who do not like the taste of cilantro.

n=15 trials x=6 successes p=.10 probability of not liking cilantro

In a recent survey of 150 teenagers, 93 stated that they always wear their seat belt in a car. find the point estimate and standard error for the proportion of teenagers that always wear a seat belt

p'=.62 σp'=.04 p'=x/n or 93/150=.62 σp'=square root of p'(1-p')/n or square root of .62(1-.62)/150= .04

In a survey, a random sample of teenagers were asked whether they have a sibling. The resulting confidence interval for the proportion of teenagers who have a sibling is (0.61,0.91). What is the sample proportion, p^?

p^=.76 A confidence interval has the form (point estimate−margin of error, point estimate+margin of error), add these numbers and divide by 2. 0.61+0.91= 1.52/2=0.76

The pizza cooking times for a local Italian restaurant follow a normal distribution with a population standard deviation of 1.2 minutes and an unknown population mean. If a random sample of 29 pizzas is taken and results in a sample mean of 8.2 minutes, find a 95% confidence interval for the population mean.

sample mean x=8.2 standrard deviation=1.2 sample size n=29 confidence level=.95, plug these into given graph to get confidence interval

Which of the following tables shows a valid probability density function? Select all correct answers.

select any charts where the values in the P(X=x) add to 1

Timothy wants to estimate the mean number of siblings for each student in his school. He records the number of siblings for each of 75 randomly selected students in the school. What is the statistic?

the mean number of siblings for the randomly selected students

A data set is summarized in the frequency table below. The data set contains a total of 30 data values. What is the missing frequency? value frequency 1 5 2 3 3 4 4 6 5 ? 6 2 7 3 8 4

the missing value is 3. add up all the frequency numbers and subtract from given number (30)

When considering different sampling methods, cluster sampling includes the steps: _______.

use simple random sampling to select a set of groups; every individual in the chosen groups is included in the sample

Suppose the length, in words, of the essays written for a contest are normally distributed and have a known population standard deviation of 325 words and an unknown population mean. A random sample of 25 essays is taken and gives a sample mean of 1640 words. Identify the parameters needed to calculate a confidence interval at the 98% confidence level. Then find the confidence interval.

x=1640 σ=325 n=25 zvalue/2=2.326 (1488.81, 1791.19) moe=151.190 so, (1640-151.19, 1640+151.19)

According to the latest financial reports from a sporting goods store, the mean sales per customer was $75 with a population standard deviation of $6. The store manager believes 39 randomly selected customers spent more per transaction.What is the probability that the sample mean of sales per customer is between $76 and $77 dollars?

μx¯=$75 σx¯=$.96 P(≤x≤77)=.13 the standard deviation is 6/√39≈$0.96. The zscores were calculated using the z score formula z1=77-75/.96=2.08 z2=76-75/.96=1.04 these numbers were found on the given table then do z1-z2=.130 so the answer is .13

A group of friends has gotten very competitive with their board game nights. They have found that overall, they each have won an average of 18 games, with a population standard deviation of 6 games. If a sample of only 2 friends is selected at random from the group, select the expected mean and the standard deviation of the sampling distribution from the options below.

σx¯=4 games μx¯=18 games The standard deviation of the sampling distribution σx¯=σ/√n=6√2≈4 games. When the distribution is normal, the mean of the sampling distribution is equal to the mean of the population μx¯=μ=18 games.


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