Concept explainers
) The housing market has recovered slowly from the economic crisis of 2008. Recently, in one
large community, realtors randomly sampled 45 bids from potential buyers to estimate the
average loss in home value. The sample showed the average loss was $10,560 with a
standard deviation of $1800.
a) Find a 95% confidence interval for the mean loss in value per home. (Use your calculator
to create this interval. Do not do the calculations by hand.)
b) Interpret this interval.
c) A neighborhood realtor predicted the average loss in value would be $11,000. Based on
your interval in part b, is the realtor right or wrong?
d) Suppose the standard deviation of the losses had been $3600 instead of $1800. What
would the larger standard deviation do to the width of the confidence interval (assuming
the same level of confidence)
Trending nowThis is a popular solution!
Step by stepSolved in 2 steps with 1 images
Suppose the standard deviation of the losses had been $3600 instead of $1800. What would the larger standard deviation do to the width of the confidence interval assuming the same level of confidence.
Suppose the standard deviation of the losses had been $3600 instead of $1800. What would the larger standard deviation do to the width of the confidence interval assuming the same level of confidence.
- Part b ?arrow_forwardYou are a researcher studying a certain species of fish in a local lake. You gather a random sample of 19 fish of this species and record the lengths (in centimeters) of the fish. The data is displayed in the table below. Estimate the mean length of all fish of this type in the lake. Use a 99% confidence level. 9.1 9 8.6 5.8 9.8 7.4 9.1 9.5 8.2 8.9 9.3 7.8 9 9.6 10.4 13.6 7.2 10.1 8.9 a) Find the confidence interval: b) What is the margin of error of your interval? c) Interpret your 99% confidence interval in contextarrow_forwardTwo friends, Karen and Jodi, work different shifts for the same ambulance service. They wonder if the different shifts average different numbers of calls. Looking at past records, Karen determines from a random sample of 32 shifts that she had a mean of 4.8 calls per shift. She knows that the population standard deviation for her shift is 1.4 calls. Jodi calculates from a random sample of 40 shifts that her mean was 4.2 calls per shift. She knows that the population standard deviation for her shift is 1.1 calls. Test the claim that there is a difference between the mean numbers of calls for the two shifts at the 0.01 level of significance. Let Karen's shifts be Population 1 and let Jodi's shifts be Population 2. Step 2 of 3 : Compute the value of the test statistic. Round your answer to two decimal places.arrow_forward
- Please help!arrow_forwardPlease answer the attached question. Thanks.arrow_forwardAccording to recent data by the American Marketing Association, 76.5%of people in the 18-24 age group used social media in 2017. You want to find out what the social media proportion is among your classmates who are all between 18 and 24 years old. Suppose you obtain a random sample of 229 students and find that 202 students in this sample are active social media users. Complete parts a through c below. a. Construct a 99% confidence interval to estimate the actual proportion of social media users. A 99% confidence interval to estimate the actual proportion has a lower limit of------ and an upper limit of------- (Round to three decimal places as needed.) b. What is the margin of error for this sample? c. Is there evidence that this proportion matches that of the overall population based on the sample?arrow_forward
- 2. According to the website Rotten Tomatoes, 94% of people who watched the Netflix show Stranger Things liked it. Since UR students tend to be scientifically minded, we hypothesize that less than 94% of UR students like the show. Out of 219 surveyed UR students, 198 indicated they like Stranger Things. Use the built in prop.test() function to: a. Construct a 90% confidence interval for the proportion of UR students who like Stranger Things. b. Test at the a=0.10 significance level whether the proportion of students who like Stranger Things is LESS THAN 0.94. What conclusion do you reach?arrow_forwardWhen you hold the mouse pointer over a point on the graph, the specific value corresponding to that point appears (e.g., “California, 2013 - 19.4%”) as well as a line called “90% Confidence” followed by an range of values. Generally speaking, what does the range of values denoted by “90% Confidence” indicate?arrow_forwardA successful basketball player has a height of 6 feet 11 inches, or 211 cm. Based on statistics from a data set, his height converts to the z score of 5.17. How many standard deviations is his height above the mean?arrow_forward
- Twelve percent of the world's population is left-handed. A researcher claims that left-handed individuals are more creative than those who aren't left-handed. In order to support his claim, he takes a sample of 1650 artists and finds that 14% of.them are left-handed. Español Answer the following. (If necessary, consult a list of formulas.) (a) Find the mean of p, where p is the proportion of left-handed individuals in a random sample of 1650. 00 (b) Find the standard deviation of p. (c) Compute an approximation for P (p s0.14), which is the probability that there will be 14% or fewer left-handed people in a random sample of 1650. Round your answer to four decimal places. Save For Later Submit Assignment Check Terms of Use Privacy Center Accessibility acer esc backspaarrow_forwardThe picture is attached. If it is not clear, the shaded region of the graph represents 5%. Thanks in advance.arrow_forwardTwo friends, Karen and Jodi, work different shifts for the same ambulance service. They wonder if the different shifts average different numbers of calls. Looking at past records, Karen determines from a random sample of 40 shifts that she had a mean of 5.1 calls per shift. She knows that the population standard deviation for her shift is 1.2 calls. Jodi calculates from a random sample of 31 shifts that her mean was 5.8 calls per shift. She knows that the population standard deviation for her shift is 1.4 calls. Test the claim that there is a difference between the mean numbers of calls for the two shifts at the 0.01 level of significance. Let Karen's shifts be Population 1 and let Jodi's shifts be Population 2. Step 3 of 3 : Draw a conclusion and interpret the decision.arrow_forward
- MATLAB: An Introduction with ApplicationsStatisticsISBN:9781119256830Author:Amos GilatPublisher:John Wiley & Sons IncProbability and Statistics for Engineering and th...StatisticsISBN:9781305251809Author:Jay L. DevorePublisher:Cengage LearningStatistics for The Behavioral Sciences (MindTap C...StatisticsISBN:9781305504912Author:Frederick J Gravetter, Larry B. WallnauPublisher:Cengage Learning
- Elementary Statistics: Picturing the World (7th E...StatisticsISBN:9780134683416Author:Ron Larson, Betsy FarberPublisher:PEARSONThe Basic Practice of StatisticsStatisticsISBN:9781319042578Author:David S. Moore, William I. Notz, Michael A. FlignerPublisher:W. H. FreemanIntroduction to the Practice of StatisticsStatisticsISBN:9781319013387Author:David S. Moore, George P. McCabe, Bruce A. CraigPublisher:W. H. Freeman