Stats Final - ch 1-12 practice test
Suppose a life insurance company sells a $220,000 one-year term life insurance policy to a 19-year-old female for $290. The probability that the female survives the year is 0.999589. Compute and interpret the expected value of this policy to the insurance company. Which of the following interpretation of the expected value is correct?
- $199.58. To find above. Step 1: 1-0.999589 = 0.000411. Step 2: $290 - 120,000 = -$219710. Step 3: Multiply $290 x 0.999589 = $289.88. Step 4: -$219,710 x 0.000411 = -$90.30. Step 5: use numbers from step 3 & 4. $289.88 + (-90.30) = $199.58. - The insurance company expects to make an average profit of $199.58 on every 19-year-old female it insures for 1 year.
Determine the required value of the missing probability to make the distribution a discrete probability distribution. P(4) = ?
- 0.26. add up everything under P(x). Then do 1 - total of P(x) = Answer.
Suppose that events E and F are independent, P(E)=0.7, and P(F)=0.9. What is the P(E and F)?
- 0.63. You multiple P(E) & P(F). 0.7 x 0.9 = 0.63.
Explain what a P-value is. What is the criterion for rejecting the null hypothesis using the P-value approach? What is the criterion for rejecting the null hypothesis using the P-value approach? Choose the correct answer below.
- A P-value is the probability of observing a sample statistic as extreme or more extreme than the one observed under the assumption that the statement in the null hypothesis is true. - If P-value<a, reject the null hypothesis.
In a survey conducted by the Gallup Organization, 1100 adult Americans were asked how many hours they worked in the previous week. Based on the results, a 95% confidence interval for the mean number of hours worked had a lower bound of 42.7 and an upper bound of 44.5. Provide two recommendations for decreasing the margin of error of the interval.
- Decrease the confidence level. - Increase the sample size.
The notation P(F E) means the probability of event ____ given event _____
- F & E.
A probability experiment is conducted in which the sample space of the experiment is S={1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}, event F={5, 6, 7, 8, 9, 10}, and event G={9, 10, 11, 12}. Assume that each outcome is equally likely. List the outcomes in F or G. Find P(F or G) by counting the number of outcomes in F or G. Determine P(F or G) using the general addition rule. List the outcomes in F or G. Select the correct choice below and, if necessary, fill in the answer box to complete your choice. Find P(F or G) by counting the number of outcomes in F or G. Determine P(F or G) using the general addition rule. Select the correct choice below and fill in any answer boxes within your choice.
- F or G = [5,6,7,8,9,10,11,12] - P(F or G) = 0.667. Get this by dividing the number from the above (8 count them above) against the total number 12 - 8/12. Round 0.6666667 = 0.667 - P(F or G) = 0.5 + 0.333 - 0.167 = 0.666. You get 0.5 by counting the numbers in F in this case was 6 then divide by 12 (total) - 6/12. You 0.333 by counting the numbers in G in this case was 4 then divide by 12 (total) - 4/12 0.666 you get by counting the outcomes (numbers in common in F & G) in this was 2, then divide by 12 - 2/12.
A manufacturer of colored candies states that 13% of the candies in a bag should be brown, 14% yellow, 13% red, 24% blue, 20% orange, and 16% green. A student randomly selected a bag of colored candies. He counted the number of candies of each color and obtained the results shown in the table. Test whether the bag of colored candies follows the distribution stated above at the α=0.05 level of significance. Determine the null and alternative hypotheses. Choose the correct answer below. Compute the expected counts for each color. What is the test statistic? Based on the results, do the colors follow the same distribution as stated in the problem?
- H0: The distribution of colors is the same as stated by the manufacturer. - To compute the expected counts use Stat > Goodness of fit > Chi square test > under observed select "frequency" > under expected select Claimed Proportion > compute. - Test statistic: 16.624 This is given along with the results from above expected counts. Will be under "chi square" right next to P value. image attached. Pvalue: 0.005 same provided w above / image attached. - Reject H0 - There is sufficient evidence that the distribution of colors is not the same as stated by the manufacturer. since p value is low.
A researcher wants to show the mean from population 1 is less than the mean from population 2 in matched-pairs data. If the observations from sample 1 are Xi and the observations from sample 2 are Yi,and di=Xi−Yi, then the null hypothesis is H0: μd=0 and the alternative hypothesis is H1: μd___ 0.
- H1: μd < 0. we know this because in the sentence it's mentioned it's LESS than.
Fill in the blank. Two events E and F are ________ if the occurrence of event E in a probability experiment does not affect the probability of event F.
- Independent.
The data on the right represent the number of traffic fatalities by seat location and gender. Determine P(male) and P(male|driver). Are the events "male" and "driver" independent? Determine P(male). Determine P(male|driver) Are the events "male" and "driver" independent?
- P(male) = 0.315 divide the total under males with the overall total. (18,141 / 57675) - P(male|driver)= 0.265 divide the driver totals, male & total. (11,885 / 44,837). - No. The occurrence of the event "driver" affects the probability of the event "male."
Determine whether the following sampling is dependent or independent. Indicate whether the response variable is qualitative or quantitative. A researcher wishes to compare academic aptitudes of patients with and without epilepsy. She obtains a random sample of 200 patients of each category who take an academic aptitude test and determines each patient's
- The sampling is independent because an individual selected for one sample does not dictate which individual is to be in the second sample. - The variable is qualitative because it classifies the individual.
Determine whether the following sampling is dependent or independent. Indicate whether the response variable is qualitative or quantitative. A researcher wishes to compare annual salaries of patients with and without narcolepsy. She obtains a random sample of 319 patients of each category who work and determines each patient's
- The sampling is independent because an individual selected for one sample does not dictate which individual is to be in the second sample. - The variable is quantitative because it is a numerical measure.
In a survey, respondents were asked to disclose their political affiliation (Democrat, Independent, Republican) and also answer the question "Would you be willing to pay higher taxes if the tax revenue went directly toward deficit reduction?" Create a contingency table and determine whether the results suggest there is an association between political affiliation and willingness to pay higher taxes to directly reduce the federal debt. Use the α=0.05 level of significance. Construct a contingency table. What are the null and alternative hypotheses for this test? Determine the test statistic. Determine the P-value. Determine the proper conclusion.
- Use stat to contruct a contingency table, just count per party who said yes / no. - H0: Political Affiliation and Taxation Willingness are independent. H1:Political Affiliation and Taxation Willingness are not independent. - t statistic: 3.864. Use Stat > Tables > Contingency > w summary> Select you're columns of Yes & no (you had to put these in separately attached image.) > Row labels in this case was "party" > Display select expected count > compute. P value - 0.145 found w above. - Do not reject H0. The data from these samples provide insufficient evidence to conclude that political affiliation and taxation willingness are not independent.
The data represent the age of world leaders on their day of inauguration. Find the five-number summary, and construct a boxplot for the data. Comment on the shape of the distribution. The five-number summary is: Choose the correct boxplot of the data below. Choose the correct description of the shape of the distribution.
- five # summary would be - MIN, Q1, MEDIAN, Q3, MAX. Open in statcrunch use stat> summary stats> colums (above). - create a box plot using statcrunch and choose the graph that's similar. - The distribution is skewed to the right.
Find the probability P(Ec)if P(E)=0.38.
= 0.62 Do this to get above: 1 - 0.38
Describe how the value of n affects the shape of the binomial probability histogram.
As n increases, the binomial distribution becomes more bell shaped.
The mean finish time for a yearly amateur auto race was 185.89 minutes with a standard deviation of 0.382 minute. The winning car, driven by Chris, finished in 185.18 minutes. The previous year's race had a mean finishing time of 110.2 with a standard deviation of 0.114 minute. The winning car that year, driven by Nina, finished in 109.84 minutes. Find their respective z-scores. Who had the more convincing victory? Which driver had a more convincing victory?
Chris' z-score: -1.86 Nina's z-score: -3.16 - Nina had a more convincing victory because of a lowerz-score.
When testing a hypothesis using the P-value Approach, if the P-value is large, reject the null hypothesis
False.
If we do not reject the null hypothesis when the statement in the alternative hypothesis is true, we have made a Type _______ error
II error.
Determine the level of measurement of the variable. Monthly temperatures: 60 F, 65 F, 70 F, 75 F, and 80 F.
Interval
A simple random sample of size n=40 is drawn from a population. The sample mean is found to be x=121.2 and the sample standard deviation is found to be s=12.3. Construct a 99% confidence interval for the population mean.
Lower bound: 115.93. Upper: 126.47 Found using T-stats, enter mean, std dev, sample size & confidenc interval > compute.
Determine the level of measurement of the variable. States in a region
Nominal
Determine the level of measurement of the variable. Positions of runners in a race
Ordinal
Determine the level of measurement of the variable. The rankings of songs in the top 100.
Ordinal
A binomial probability experiment is conducted with the given parameters. Compute the probability of x successes in the n independent trials of the experiment. n = 7, p = 0.9, x = 4. P(4) =
P(4) = 0.0230 rounded 4 dec. Use Stat > Cal> Binomial > enter n & p and then in the P(x = 4) Compute > attached image.
Let the sample space be S={1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. Suppose the outcomes are equally likely. Compute the probability of the event E="an odd number less than 8." P(E) =
P(E) = 0.4 Count the amount of numbers that are odd below 8 and then divide against the entire number. (4/10).
Suppose that E and F are two events and that P(E and F)=0.3 and P(E)=0.4. What is P(F|E)?
P(F|E) = 0.75. Divide 0.3 / 0.4.
Determine the point estimate of the population mean and margin of error for the confidence interval. Lower bound is 18, upper bound is 28. The point estimate of the population mean is _____ The margin of error for the confidence interval is _____
Point Estimate: 23. Use desmos 18+28 / 2. Margin of error: 5. Use demos 18-28 / 2 = -5, *remove negative when entering answer*.
Microsoft wants to administer a satisfaction survey to its current customers. Using their customerdatabase, the company randomly selects 70 customers and asks them about their level of satisfaction with the company. What type of sampling is used?
Simple random
Explain what "statistical significance" means.
Statistical significance means that the result observed in a sample is unusual when the null hypothesis is assumed to be true.
Define statistics.
Statistics is the science of collecting, organizing, summarizing, and analyzing information to draw a conclusion and answer questions. In addition, statistics is about providing a measure of confidence in any conclusions.
A histogram of a set of data indicates that the distribution of the data is skewed right. Which measure of central tendency will likely be larger, the mean or the median? Why?
The mean will likely be larger because the extreme values in the right tail tend to pull the mean in the direction of the tail.
Mark performed a two-sample z-test for proportions to test the hypothesis that there was no difference in the proportion who support increasing student fees between male and female students at a particular university. Mark obtained a z-statistic of 0. Based on this information, which of the following is always true?
The sample proportions will be the same for both male and female students at this university.
Determine whether the study depicts an observational study or an experiment. Professional weightlifters are randomly divided into two groups. One group is given a diet of mostly protein; the other is given a diet of mostly carbohydrates. After 1 week, each group is given a lifting test to compare strength.
The study is an experiment because the researchers control one variable to determine the effect on the response variable.
An advertisement for a diet program described a study performed on 70 obese adults. Each person in the study was weighed before beginning the diet and then 5 weeks after starting the diet. The difference was recorded-a positive value indicates a person lost weight on the diet while a negative value indicates the person gained weight while on the diet. The advertisement gave a 95% confidence interval for the average weight change while on the diet: (−1,5)lbs. The advertisement says that this shows the diet works at reducing weight for obese adults since more people lost weight than gained weight. What conclusion can be made about the weight loss program?
Their interpretation of the confidence interval is not correct. Because 0 is in the interval, there is no evidence to indicate that the average weight loss is more than 0 for obese adults.
A certain marathon has had a wheelchair division since 1977. An interested fan wondered who is faster: the men's marathon winner or the women's wheelchair marathon winner, on average. A paired t-test was performed, and the p-value was found to be 0.001. Which of the following is the correct conclusion?
There is sufficient evidence to indicate that the men's running winning time and the women's wheelchair winning time each year are different, on average.
The following data represent exam scores in a statistics class taught using traditional lecture and a class taught using a "flipped" classroom. Complete parts (a) through (c) below. (a) Which course has more dispersion in exam scores using the range as the measure of dispersion? (b) Which course has more dispersion in exam scores using the sample standard deviation as the measure of dispersion? (c) Suppose the score of 59.5 in the traditional course was incorrectly recorded as 595. How does this affect the range? How does this affect the standard deviation? What property does this illustrate?
To find all use Stat > Summary stats > columns > select range, std dev > select both columns and compute. for last question in C - Neither the range nor the standard deviation is resistant.
The following data represent the amount of time (in minutes) a random sample of eight students took to complete the online portion of an exam in a particular statistics course. Compute the mean, median, and mode time.
Use stat crunch > Stat > Summary Stats > Columns > select column & mean mod and median.
Suppose a sample of O-rings was obtained and the wall thickness (in inches) of each was recorded. Use a normal probability plot to assess whether the sample data could have come from a population that is normally distributed.
Use stat crunch, to find the correlation. Graph QQ plot Select column Add: Correlation statistic Other options: Normal quantiles on y-axis Compute. Next compare the correlation to the critical value of the sum. Sum in this case was 16. Correlation was 0.996 Critical value of 16 was 0.941, and that's how you find the below answer. Yes. The correlation between the expected z-scores and the observeddata, . 996. exceeds the critical value, 0.941. Therefore, it is reasonable to conclude that the data come from a normal population.
Assume the random variable X is normally distributed with mean μ=50 and standard deviation σ=7. Compute the probability. Be sure to draw a normal curve with the area corresponding to the probability shaded. Which of the following normal curves corresponds to P(35<X<62)?
Use stat>cal>normal> enter mean & std dev > between > #'s > compute. choose correct graph. P(35<X<62) = 0.9407 (rounded 4 dec)
A cellular phone company monitors monthly phone usage. The following data represent the monthly phone use in minutes of one particular customer for the past 20 months. Use the given data to answer parts (a) and (b). (a) Determine the standard deviation and interquartile range of the data. s= IQR = (b) Suppose the month in which the customer used 334 minutes was not actually that customer's phone. That particular month the customer did not use their phone at all, so 0 minutes were used. How does changing the observation from 334 to 0 affect the standard deviation and interquartile range? What property does this illustrate? What property does this illustrate? Choose the correct answer below.
a) s= 74.33 IQR = 133.5 Both found using Stat crunch. b) The standard deviation increases and the interquartile range is not affected. - Resistance.
Suppose that a recent poll found that 48% of adults believe that the overall state of moral values is poor. Complete parts (a) through (c). (a) For 300 randomly selected adults, compute the mean and standard deviation of the random variable X, the number of adults who believe that the overall state of moral values is poor. (b) Interpret the mean. Choose the correct answer below. (c) Would it be unusual if 141 of the 300 adults surveyed believe that the overall state of moral values is poor?
a) -The mean of X is = 144. Use desmos - 300 x 0.48= 144. - The std dev of X is 8.7. Use desmos sqrt all 144(1-0.48) = 8.65... round > 8.7. b) For every 300 adults, the mean is the number of them that would be expected to believe that the overall state of moral values is poor. c) No
About 7% of the population of a large country is hopelessly romantic. If two people are randomly selected, what is the probability both are hopelessly romantic? What is the probability at least one is hopelessly romantic? Assume the events are independent. (a) The probability that both will be hopelessly romantic is nothing
a) 0.0049 Since they're both independent use multiplication rule. 0.07 x 0.07 = 0.0049. b) 0.1351. 1st: 1 - 0.07 = 0.93 2nd: 0.93 x 0.93 = 0.8649. 3rd: 1 - 0.8649 = 0.1351.
A baseball player hit 58 home runs in a season. Of the 58 home runs, 18 went to right field, 16 went to right center field, 11 went to center field, 11 went to left center field, and 2 went to left field. (a) What is the probability that a randomly selected home run was hit to right field? (b) What is the probability that a randomly selected home run was hit to left field? (c) Was it unusual for this player to hit a home run to left field? Explain.
a) 0.310 Divide 18 (right field) over the total home runs (58) - 18/58. b) 0.034 Divide 2 (left field) over the total home runs (58) - 2/58. c) Yes, because P(left field) < 0.05
The time required for an automotive center to complete an oil change service on an automobile approximately follows a normal distribution, with a mean of 17 minutes and a standard deviation of 3 minutes. (a) The automotive center guarantees customers that the service will take no longer than 20 minutes. If it does take longer, the customer will receive the service for half-price. What percent of customers receive the service for half-price? (b) If the automotive center does not want to give the discount to more than 2% of its customers, how long should it make the guaranteed time limit?
a) 15.87% use Stat>Cal>Normal In P enter > 20 then compute. b) 24 Use Stat > cal > normal > this time in the equal type 0.02 (whatever percent) gave 23.16. Had to round to 24.
The mean incubation time of fertilized eggs is 20 days. Suppose the incubation times are approximately normally distributed with a standard deviation of 1 day. (a) Determine the 14th percentile for incubation times. (b) Determine the incubation times that make up the middle 39% of fertilized eggs.
a) 19. Use Stat> Cal > Normal > Enter mean & std dev provided > in P enter < (less than) and in the = enter 0.14. it will give the answer in P (x < 18.919) - rounded to 19. b) 19 to 21 (rounded to one number) Same as above, but enter = 0.39 to compute the between.
The following frequency histogram represents the IQ scores of a random sample of seventh-grade students. IQs are measured to the nearest whole number. The frequency of each class is labeled above each rectangle. Use the histogram to answers parts (a) through (g). (a) How many students were sampled? (b) Determine the class width. (c) Identify the classes and their frequencies. Choose the correct answer below. (d) Which class has the highest frequency? (e) Which class has the lowest frequency? (f) What percent of students had an IQ of at least 120? (g) Did any students have an IQ of 163?
a) 200. add up all the numbers in the graph. b) 10 notice it goes from 90 - 100, so the width is 10. c) 60-69, 1; 70-79, 2; 80-89, 13; 90-99, 44; 100-109, 53; 110-119, 41; 120-129, 31; 130-139, 9; 140-149, 4; 150-159, d) 100 - 109. you can find by looking at the graph. e) 60 -69 you can find by looking at the graph. f) 23%. use graph add up anything after 120, divide it by the total number and turn into a percent. g) No, because there are no bars, or frequencies, greater than an IQ of 160.
According to an airline, flights on a certain route are on time 75% of the time. Suppose 13 flights are randomly selected and the number of on-time flights is recorded. (a) Explain why this is a binomial experiment. (b) Find and interpret the probability that exactly 8 flights are on time. (c) Find and interpret the probability that fewer than 8 flights are on time. (d) Find and interpret the probability that at least 8 flights are on time. (e) Find and interpret the probability that between 6 and 8 flights, inclusive, are on time. (a) Identify the statements that explain why this is a binomial experiment. Select all that apply.
a) Answers: - The experiment is performed a fixed number of times. - The trials are independent. - The probability of success is the same for each trial of the experiment. - There are two mutually exclusive outcomes, success or failure. b) 0.1258 Use Stat > Cal > Binomial > enter n = 13 & p = 0.75 and then the > or < or = etc (depends on your order). > compute. Interpretation: In 100 trials of this experiment, it is expected about 13 to result in exactly 8 flights being on time. c) 0.0802 same stat steps as above. Interpretation: In 100 trials of this experiment, it is expected about 8 to result in exactly 8 flights being on time. d) 0.9198 same stat steps as in pt B. Interpretation: In 100 trials of this experiment, it is expected about 92 to result in exactly 8 flights being on time. e) 0.2004 same stat steps as in pt B. Interpretation: In 100 trials of this experiment, it is expected about 20 to result in exactly 8 flights being on time.
Three years ago, the mean price of an existing single-family home was $243,754. A real estate broker believes that existing home prices in her neighborhood are higher. (a) Determine the null and alternative hypotheses. (b) Explain what it would mean to make a Type I error. (c) Explain what it would mean to make a Type II error.
a) H0: mean = $243,754 H1: mean > (greater than) $243,754. We know it's greater than because the sentence says "higher." b) The broker rejects the hypothesis that the mean price is $243,754, when it is the true mean cost. c) The broker fails to reject the hypothesis that the mean price is $243,754, when the true mean price is greater than $243,754.
Conduct a test at the α=0.01 level of significance by determining (a) the null and alternative hypotheses, (b) the test statistic, and (c) the P-value. Assume the samples were obtained independently from a large population using simple random sampling. Test whether p1>p2. The sample data are x1=120, n1=248, x2=137, and n2=319. (a) Choose the correct null and alternative hypotheses below. (b) Determine the test statistic. (c) Determine the P-value What is the result of this hypothesis test?
a) H0:p1 = p2 versus H1: p1>p2. b) 1.30 rounded two dec. Stat > Proportion Stats > two samples> w summary > type in the Ho p1...etc > and confidence interval > compute. - 0.098 rounded 3 dec. Got with pt B stat steps. -Do not reject the null hypothesis because there is not sufficient evidence to conclude that p1>p2.
Several years ago, the mean height of women 20 years of age or older was 63.7 inches. Suppose that a random sample of 45 women who are 20 years of age or older today results in a mean height of 65.1 inches. (a) State the appropriate null and alternative hypotheses to assess whether women are taller today. (b) Suppose the P-value for this test is 0.09. Explain what this value represents. (c) Write a conclusion for this hypothesis test assuming an α=0.10 level of significance.
a) H0:μ=63.7 in. versus H1: μ>63.7 in. b) There is a 0.09 probability of obtaining a sample mean height of 65.1 inches or taller from a population whose mean height is 63.7 inches. we know it's "or taller" because in A we chose "greater than" symbol. c) Reject the null hypothesis. There is sufficient evidence to conclude that the mean height of women 20 years of age or older is greater today.
The data in the following table show the association between cigar smoking and death from cancer for 140,684 men. Note: current cigar smoker means cigar smoker at time of death. (a) If an individual is randomly selected from this study, what is the probability that he died from cancer? (b) If an individual is randomly selected from this study, what is the probability that he was a current cigar smoker? (c) If an individual is randomly selected from this study, what is the probability that he died from cancer and was a current cigar smoker? (d) If an individual is randomly selected from this study, what is the probability that he died from cancer or was a current cigar smoker?
a) P(died from cancer) = 0.008. You get this by adding up everything under "died from cancer" in image attached an dividing it over the total. - 1166/ 140684. Round 3 dec. b)P(current cigar smoker) = 0.041 you get the above by adding *both* totals under current cigar smoker the ones what died w cancer and w/o cancer, then dividing by the total: 171 + 5624 = 5795. Then 5795 / 140684. c) 0.001 In this case we did 171 (current cigar smokers & died w cancer)/ 140684 (total). d) Add pt A + B - C = 0.008 + 0.041 - 0.001 = 0.048.
The following data represent the number of games played in each series of an annual tournament from 1932 to 2004. Complete parts (a) through (d) below. (a) Construct a discrete probability distribution for the random variable x. (b) Graph the discrete probability distribution. Choose the correct graph below. (c) Compute and interpret the mean of the random variable x. Interpret the mean of the random variable x. (d) Compute the standard deviation of the random variable x
a) P(x) 0.2222, 0.1667, 0.2083 & 0.4028. To find add everything under "frequency" then divide each frequency against that total ex. 16 / 72 - 0.2222. b) Use statcrunch Stat > Cal> Custom > select X & Y > compute. Choose which best looks like the graph you made. c) 5.7917 - When you do pt B, that same graph provides you with the mean. Do NOT go to summary stats to get a mean, get directly from the graph in pt b. - Interpretation: The series, if played many times, would be expected to last about 5.8 games, on average. d) 1.2 rounded one dec. Found in the graph from pt B.
The data shown to the right represent the age (in weeks) at which babies first crawl, based on a survey of 12 mothers. Complete parts (a) through (c) below. (a) Draw a normal probability plot to determine if it is reasonable to conclude the data come from a population that is normally distributed. Choose the correct answer below. Is it reasonable to conclude that the data come from a population that is normally distributed? (b) Draw a boxplot to check for outliers. Choose the correct answer below. Does the boxplot suggest that there are outliers? (c) Construct and interpret a 95% confidence interval for the mean age at which a baby first crawls. Select the correct choice and fill in the answer boxes to complete your choice.
a) QQ Plot. in Stat, then choose which one compares the best. Also, confirm if it's linear in this case it was so - Yes, because the plotted values are approximately linear. b) Draw box plot, choose which best compares and check if any outliers. This case no outliers so: No, there are no points that are outside of the 1.5(IQR) boundary. c) The lower bound is 33.8 weeks and the upper bound is 45 weeks. We are 95% confident that the mean age at which a baby first crawls is within the confidence interval. Found this by computing the mean & std first then opening T stats, entering mean, std dev and sample size, adjust confidence interval if needed > compute.
A simple random sample of size n=48 is obtained from a population with μ=64 and σ=15. (a) What must be true regarding the distribution of the population in order to use the normal model to compute probabilities involving the sample mean? Assuming that this condition is true, describe the sampling distribution of x. (b) Assuming the normal model can be used, determine P(x<67.7). (c) Assuming the normal model can be used, determine P(x≥66.3). (a) What must be true regarding the distribution of the population?
a) Since the sample size is large enough, the population distribution does not need to be normal. b) P(x<67.7)= 0.9563. Use stat>cal>normal> enter mean>std dev /sqrt n > P > compute. c)P(x≥66.3)= 0.1440 stat>cal>normal> enter mean>std dev > P > compute.
The data to the right represent the number of customers waiting for a table at 6:00 P.M. for 40 consecutive Saturdays at Bobak's Restaurant. Complete parts (a) through (h) below. (a) Are these data discrete or continuous? Explain. (b) Construct a frequency distribution of the data. (c) Construct a relative frequency distribution of the data. (d) What percentage of the Saturdays had 4 or more customers waiting for a table at 6:00 P.M.? (e) What percentage of the Saturdays had 12 or fewer customers waiting for a table at 6:00 P.M.? (f) Construct a frequency histogram of the data. Choose the correct histogram below. (g) Construct a relative frequency histogram of the data. Choose the correct histogram below. h) Describe the shape of the distribution. Choose the correct answer below.
a) The data are discrete because there are a finite or countable number of values. b) Use Stat > Tables > Frequency > select column > choose statistic: frequency > compute. Add up 4-6 etc. c) Use Stat > Tables > Frequency > select column > choose statistic: relative frequency > compute. Add up 4-6 etc. d) 97.5% add up the percentages in pt C that had 4 or more. Then move dec one to the right. e) 97.5% add up the percentages in pt C of 12 or fewer. f) Graph > Histogram > select column > frequency > compute. Choose the graph that looks similar. g)Graph > Histogram > select column > relative frequency > compute. Choose the graph that looks similar. h)The distribution is symmetric because the left and right sides are approximately mirror images
Determine whether the random variable is discrete or continuous. In each case, state the possible values of the random variable. (a) The number of fish caught during a fishing tournament. (b) The weight of a T-bone steak. (a) Is the number of fish caught during a fishing tournament discrete or continuous? (b) Is the weight of a T-bone steak discrete or continuous?
a) The random variable is discrete. The possible values are x=0,1, 2,... b) The random variable is continuous. The possible values are w>0.
The shape of the distribution of the time required to get an oil change at a 10-minute oil-change facility is unknown. However, records indicate that the mean time is 11.2 minutes, and the standard deviation is 3.1 minutes. Complete parts (a) through (c). (a) To compute probabilities regarding the sample mean using the normal model, what size sample would be required? (b) What is the probability that a random sample of n=35 oil changes results in a sample mean time less than 10 minutes? The probability is approximately ____ (c) Suppose the manager agrees to pay each employee a $50 bonus if they meet a certain goal. On a typical Saturday, the oil-change facility will perform 35 oil changes between 10 A.M. and 12 P.M. Treating this as a random sample, there would be a 10% chance of the mean oil-change time being at or below what value? This will be the goal established by the manager.
a) The sample size needs to be greater than or equal to 30. b) 0.0110 - use stat> cal > normal> mean > std/sqrt n > less than 10 > compute. c) 10.5 Image attached on how to get this.
A researcher studies water clarity at the same location in a lake on the same dates during the course of a year and repeats the measurements on the same dates 5 years later. The researcher immerses a weighted disk painted black and white and measures the depth (in inches) at which it is no longer visible. The collected data is given in the table below. Complete parts (a) through (c) below. a) Why is it important to take the measurements on the same date? b) Does the evidence suggest that the clarity of the lake is improving at the α=0.05 level of significance? Note that the normal probability plot and boxplot of the data indicate that the differences are approximately normally distributed with no outliers. Let di=Xi−Yi. Identify the null and alternative hypotheses. Determine the test statistic for this hypothesis test. Find the P-value for this hypothesis test. What is your conclusion regarding H0? c) Draw a boxplot of the differenced data. Does this visual evidence support the results obtained in part b)? Does this visual evidence support the results obtained in part b)?
a) Using the same dates makes the second sample dependent on the first and reduces variability in water clarity attributable to date. b) H0: mean d = 0 h1: mean d < 0 Keep it at 0. Test statistic: 0.77. Found using Stat> T Stat> Paired > Enter the H0 & H1 from above > sample 1 & 2 (x1 / x2) > compute. Keep confidence interval as is. P value: 0.763 Found w above. -Do not reject H0. There is not sufficient evidence at the α=0.05 level of significance to conclude that the clarity of the lake is improving. c) To box plot the difference, first do x1 - x2, making a separate column in stat crunch and box plotting only that column. Yes because the boxplot supports that the lake is not becoming more clear, since most differences are positive or near 0.
Ramp metering is a traffic engineering idea that requires cars entering a freeway to stop for a certain period of time before joining the traffic flow. The theory is that ramp metering controls the number of cars on the freeway and the number of cars accessing the freeway, resulting in a freer flow of cars, which ultimately results in faster travel times. To test whether ramp metering is effective in reducing travel times, engineers conducted an experiment in which a section of freeway had ramp meters installed on the on-ramps. The response variable for the study was speed of the vehicles. A random sample of 15 cars on the highway for a Monday at 6 p.m. with the ramp meters on and a second random sample of 15 cars on a different Monday at 6 p.m. with the meters off resulted in the following speeds (in miles per hour). a) Draw side-by-side boxplots of each data set. Does there appear to be a difference in the speeds? Are there any outliers? Choose the correct box plot below. Does there appear to be a difference in the speeds? Are there any outliers? (b) Are the ramp meters effective in maintaining a higher speed on the freeway? Use the α=0.01 level of significance. State the null and alternative hypotheses. Choose the correct answer below. Determine the P-value for this test. Choose the correct conclusion.
a) draw boxplots and choose which best matches. -Yes, the Meters On data appear to have higher speeds. *you get this by looking at the box plots.* - No, there does not appear to be any outliers. b) attached image. (mean off) P value: 0.047 Find using T Stats> Two samples> but first open the summary in stat crunch compute mean & std dev for each sample. - Do not reject H0. There is not sufficient evidence at the α=0.01 level of significance that the ramp meters are effective in maintaining higher speed on the freeway.
Two researchers conducted a study in which two groups of students were asked to answer 42 trivia questions from a board game. The students in group 1 were asked to spend 5 minutes thinking about what it would mean to be a professor, while the students in group 2 were asked to think about soccer hooligans. These pretest thoughts are a form of priming. The 200 students in group 1 had a mean score of 25.3 with a standard deviation of 4.8, while the 200 students in group 2 had a mean score of 18.2 with a standard deviation of 4.3. Complete parts (a) and (b) below. (a) Determine the 90% confidence interval for the difference in scores, μ1−μ2. Interpret the interval. Interpret the interval. Choose the correct answer below. (b) What does this say about priming?
a) lower: 6.349 upper: 7.851 Use T Stat > Two sample > Enter two samples> leave H0 as is > enter confidence interval > compute. - The researchers are 90% confident that the difference of the means is in the interval b) Since the 90% confidence interval does not contain zero, the results suggest that priming does have an effect on scores
According to a survey in a country, 39% of adults do not own a credit card. Suppose a simple random sample of 600 adults is obtained. Complete parts (a) through (d) below. (a) Describe the sampling distribution of p hat,the sample proportion of adults who do not own a credit card. Choose the phrase that best describes the shape of the sampling distribution of p hat below. Determine the mean of the sampling distribution of p. (b) What is the probability that in a random sample of 600 adults, more than 43% do not own a credit card? c) What is the probability that in a random sample of 600 adults, between 38% and 43% do not own a credit card? d) Would it be unusual for a random sample of 600 adults to result in 228 or fewer who do not own a credit card? Why? Select the correct choice below and fill in the answer box to complete your choice.
a)Approximately normal because n≤0.05N and np(1-p)>10. Mean = 0.39 Std dev = 0.20 Get this by using desmos. Sqrt ALL 0.39 (1-0.39) / 600. Round 3 dec. b) 0.0228 Use stat normal cal. 0.43 goes in the P part. compute. If 100 different random samples of 600 adults were obtained, one would expect 2 to result in more than 43% not owning a credit card. c) 0.6687 Same stat steps above. If 100 different random samples of 600 adults were obtained, one would expect 67 to result in between 38% and 43% not owning a credit card. d)The result is not unusual because the probability that p is less than or equal to the sample proportion is 0.3078,which is greater than 5%.
Researchers initiated a long-term study of the population of American black bears. One aspect of the study was to develop a model that could be used to predict a bear's weight (since it is not practical to weigh bears in the field). One variable thought to be related to weight is the length of the bear. The accompanying data represent the lengths and weights of 12 American black bears. Complete parts (a) through (d) below. (a) Which variable is the explanatory variable based on the goals of the research? (b) Draw a scatter diagram of the data. Choose the correct graph below. (c) Determine the linear correlation coefficient between weight and length. (d) Does a linear relation exist between the weight of the bear and its length?
a)The length of the bear b) go to graph>scatter plot> choose length as X > compute. Then adjust the numbers on the x and y-axis. choose best looking graph. c) Linear correlation: 0.700. Stat>Summary stats> correlation> select both length and weight > compute. Round 3 dec. d) The variables weight of the bear and length of the bear are positively associated because r is positive and the absolute value of the correlation coefficient is, 0.700 is greater than the critical value, 0.576. the graph will let you know if its positive / negative, critical value you get from table provided.
Match the linear correlation coefficient to the scatter diagram. The scales on the x- and y-axis are the same for each scatter diagram. (a) r=−1, (b) r=−0.049, (c) r=−0.969
answer in image attached.
Compute the critical value za/2 that corresponds to a 86% level of confidence.
za/2 = 1.48 Use Stat > Cal > Normal> Leave mean & std dev as is > enter confidence level in the equal > choose between > compute. Only use the positive answer for this question.
