Concept explainers
a.
Check whether the
a.

Answer to Problem 39SE
The median sale price for single-family homes in St. Louis is less than the national median price of $180,000.
Explanation of Solution
Calculation:
The given information is that the national median sales price for single family home is $180,000. The level of significance is 0.05.
The hypotheses are given below:
Null hypothesis:
That is, the median sales price in St. Louis is significantly greater than or equal to the national median of $180,000.
Alternative hypothesis:
That is, the median sales price in St. Louis is significantly lower than the national median of $180,000.
For a sign test, the mean
The mean is,
Thus, the mean is 25.
The standard deviation is,
Thus, the standard deviation is 3.5355.
There are 18 plus signs in the lower tail.
The probability of 18 plus signs in the lower tail can be obtained by using the continuity correction factor and normal approximation. Hence, the p-value is obtained by using normal distribution with
Procedure:
Step by step procedure to obtain the above probability using Table 1 of Appendix B is given below:
- Locate the value –1.8 in the column, named z.
- Move towards the right along the row of –1.8, till the column named 0.04 is reached.
- The cell at the intersection of the row –1.8 and the column 0.04 gives the cumulative probability corresponding to the standard normal variable value –1.84.
Thus,
Now,
Rejection rule:
If
Conclusion:
Here the level of significance
Here,
That is, p-value is less than significance level.
Therefore, the null hypothesis is rejected.
Hence, it can be concluded that the median sale price for single-family homes in St. Louis is less than the national median price of $180,000.
b.
Check whether the median sales price in Denver is significantly higher than the national median of $180,000 or not.
b.

Answer to Problem 39SE
The median sale price for single-family homes in Denver is greater than the national median price of $180,000.
Explanation of Solution
Calculation:
The hypotheses are given below:
Null hypothesis:
That is, the median is less than or equal to 180,000.
Alternative hypothesis:
That is, median is greater than 180,000.
There are
For a sign test, the mean
The mean is,
Thus, the mean is 20.
The standard deviation is,
Thus, the standard deviation is 3.1623.
There 27 plus signs in the upper tail.
The probability of 27 plus signs in the upper tail can be obtained by using the continuity correction factor and normal approximation. Hence, the p-value is obtained by using normal distribution with
Procedure:
Step by step procedure to obtain the above probability using Table 1 of Appendix B is given below:
- Locate the value 2.0 in the column, named z.
- Move towards the right along the row of 2.0, till the column named 0.06 is reached.
- The cell at the intersection of the row 2.0 and the column 0.06 gives the cumulative probability corresponding to the standard normal variable value 2.06.
Thus,
Therefore,
The upper-tail p-value is 0.0197.
Rejection rule:
If
Conclusion:
Here the level of significance
Here,
That is, p-value is less than significance level.
Therefore, the null hypothesis is rejected.
Hence, it can be concluded that the median sale price for single-family homes in Denver is greater than the national median price of $180,000.
Want to see more full solutions like this?
Chapter 18 Solutions
Modern Business Statistics with Microsoft Office Excel (with XLSTAT Education Edition Printed Access Card)
- A researcher wishes to estimate, with 90% confidence, the population proportion of adults who support labeling legislation for genetically modified organisms (GMOs). Her estimate must be accurate within 4% of the true proportion. (a) No preliminary estimate is available. Find the minimum sample size needed. (b) Find the minimum sample size needed, using a prior study that found that 65% of the respondents said they support labeling legislation for GMOs. (c) Compare the results from parts (a) and (b). ... (a) What is the minimum sample size needed assuming that no prior information is available? n = (Round up to the nearest whole number as needed.)arrow_forwardThe table available below shows the costs per mile (in cents) for a sample of automobiles. At a = 0.05, can you conclude that at least one mean cost per mile is different from the others? Click on the icon to view the data table. Let Hss, HMS, HLS, Hsuv and Hмy represent the mean costs per mile for small sedans, medium sedans, large sedans, SUV 4WDs, and minivans respectively. What are the hypotheses for this test? OA. Ho: Not all the means are equal. Ha Hss HMS HLS HSUV HMV B. Ho Hss HMS HLS HSUV = μMV Ha: Hss *HMS *HLS*HSUV * HMV C. Ho Hss HMS HLS HSUV =μMV = = H: Not all the means are equal. D. Ho Hss HMS HLS HSUV HMV Ha Hss HMS HLS =HSUV = HMVarrow_forwardQuestion: A company launches two different marketing campaigns to promote the same product in two different regions. After one month, the company collects the sales data (in units sold) from both regions to compare the effectiveness of the campaigns. The company wants to determine whether there is a significant difference in the mean sales between the two regions. Perform a two sample T-test You can provide your answer by inserting a text box and the answer must include: Null hypothesis, Alternative hypothesis, Show answer (output table/summary table), and Conclusion based on the P value. (2 points = 0.5 x 4 Answers) Each of these is worth 0.5 points. However, showing the calculation is must. If calculation is missing, the whole answer won't get any credit.arrow_forward
- Binomial Prob. Question: A new teaching method claims to improve student engagement. A survey reveals that 60% of students find this method engaging. If 15 students are randomly selected, what is the probability that: a) Exactly 9 students find the method engaging?b) At least 7 students find the method engaging? (2 points = 1 x 2 answers) Provide answers in the yellow cellsarrow_forwardIn a survey of 2273 adults, 739 say they believe in UFOS. Construct a 95% confidence interval for the population proportion of adults who believe in UFOs. A 95% confidence interval for the population proportion is ( ☐, ☐ ). (Round to three decimal places as needed.)arrow_forwardFind the minimum sample size n needed to estimate μ for the given values of c, σ, and E. C=0.98, σ 6.7, and E = 2 Assume that a preliminary sample has at least 30 members. n = (Round up to the nearest whole number.)arrow_forward
- In a survey of 2193 adults in a recent year, 1233 say they have made a New Year's resolution. Construct 90% and 95% confidence intervals for the population proportion. Interpret the results and compare the widths of the confidence intervals. The 90% confidence interval for the population proportion p is (Round to three decimal places as needed.) J.D) .arrow_forwardLet p be the population proportion for the following condition. Find the point estimates for p and q. In a survey of 1143 adults from country A, 317 said that they were not confident that the food they eat in country A is safe. The point estimate for p, p, is (Round to three decimal places as needed.) ...arrow_forward(c) Because logistic regression predicts probabilities of outcomes, observations used to build a logistic regression model need not be independent. A. false: all observations must be independent B. true C. false: only observations with the same outcome need to be independent I ANSWERED: A. false: all observations must be independent. (This was marked wrong but I have no idea why. Isn't this a basic assumption of logistic regression)arrow_forward
- Business discussarrow_forwardSpam filters are built on principles similar to those used in logistic regression. We fit a probability that each message is spam or not spam. We have several variables for each email. Here are a few: to_multiple=1 if there are multiple recipients, winner=1 if the word 'winner' appears in the subject line, format=1 if the email is poorly formatted, re_subj=1 if "re" appears in the subject line. A logistic model was fit to a dataset with the following output: Estimate SE Z Pr(>|Z|) (Intercept) -0.8161 0.086 -9.4895 0 to_multiple -2.5651 0.3052 -8.4047 0 winner 1.5801 0.3156 5.0067 0 format -0.1528 0.1136 -1.3451 0.1786 re_subj -2.8401 0.363 -7.824 0 (a) Write down the model using the coefficients from the model fit.log_odds(spam) = -0.8161 + -2.5651 + to_multiple + 1.5801 winner + -0.1528 format + -2.8401 re_subj(b) Suppose we have an observation where to_multiple=0, winner=1, format=0, and re_subj=0. What is the predicted probability that this message is spam?…arrow_forwardConsider an event X comprised of three outcomes whose probabilities are 9/18, 1/18,and 6/18. Compute the probability of the complement of the event. Question content area bottom Part 1 A.1/2 B.2/18 C.16/18 D.16/3arrow_forward
- Glencoe Algebra 1, Student Edition, 9780079039897...AlgebraISBN:9780079039897Author:CarterPublisher:McGraw HillBig Ideas Math A Bridge To Success Algebra 1: Stu...AlgebraISBN:9781680331141Author:HOUGHTON MIFFLIN HARCOURTPublisher:Houghton Mifflin HarcourtHolt Mcdougal Larson Pre-algebra: Student Edition...AlgebraISBN:9780547587776Author:HOLT MCDOUGALPublisher:HOLT MCDOUGAL


