3. Consider the following sample data and hypotheses. Assume that the populations are normally distributed with unequal variances.
Sample Mean1 = 262 Sample Variance1 = 23 n1 = 10
Sample Mean2 = 249 Sample Variance2 = 35 n2 = 10
a. Construct the 90% Confidence Interval for the difference of the two means.
H0: μ1 – μ2 ≤ 0
HA: μ1 – μ2 > 0
b. Using the hypotheses listed above, conduct the following hypothesis test steps. Following the “Roadmap for Hypothesis Testing”, State Null and Alternative Hypotheses; Calculate the Test Statistic; Determine the Critical Value for α = 0.05; Draw a picture complete with Test Statistic, Critical Value & Rejection Zone; Determine the Conclusion reached by the Hypothesis Test using the Critical Value Approach.
In: Statistics and Probability
One card is drawn from an ordinary deck of 52 cards. Find the probabilities of drawing the following cards.
a. An 8 or 2
b. A black card a 9
c. A 2 or a black 7
d. A heart or a black card
e. A face card or a diamond
a. What is the probability that the card is either an 8 or a 2 ?
(Simplify your answer. Type an integer or a fraction.)
In: Statistics and Probability
Statistics concepts for engineering management:
The data in the table provides: College GPA, High School GPA,
SAT total score, and number of letters of reference.
a. Generate a model for college GPA as a function of
the other three variables.
b. Is this model useful? Justify your
conclusion.
c. Are any of the variables not useful predictors?
Why?
CGPA HSGPA SAT REF
2.04 2.01 1070 5
2.56 3.4 1254 6
3.75 3.68 1466 6
1.1 1.54 706 4
3 3.32 1160 5
0.05 0.33 756 3
1.38 0.36 1058 2
1.5 1.97 1008 7
1.38 2.03 1104 4
4.01 2.05 1200 7
1.5 2.1 896 7
1.29 1.34 848 3
1.9 1.51 958 5
3.11 3.12 1246 6
1.92 2.14 1106 4
0.81 2.6 790 5
1.01 1.9 954 4
3.66 3.06 1500 6
2 1.6 1046 5
In: Statistics and Probability
ANALYZE/INTERPRET THE STUDY AS WELL AS YOUR CONCLUSIONS FROM YOUR CALCULATIONS
DVR’s allow us to skip commercials when watching TV. Some also allow advertisers to see which ads are actually watched. Advertising research indicates that an ad has to be viewed by more than 5% of households to be cost effective (at the very least break even) in the long run. A sample of 250 homes in Cincinnati were monitored to determine viewers’ reactions to a fast food commercial that featured a well-known band. The data revealed 20 of the homes actually viewed the ad
On the basis of this study, will the ad be cost effective for the company? Use the 0.01 significance level.
In: Statistics and Probability
The table below contains the amount that a sample of nine customers spent for lunch ($) at a fast-food restaurant:
4.88 5.01 5.79 6.35 7.39 7.68 8.23 8.71 9.88
a. Compute the sample mean and sample standard deviation of the
amount spent for lunch.
b. Construct a 99% confidence interval estimate for the population mean amount spent for lunch ($) at the fast-food restaurant, assuming the population is normally distributed.
c. Interpret the interval constructed in (b).
In: Statistics and Probability
Let X be the number of students who attend counseling at a certain time of day. Suppose that the probability mass function is as follows p (0) = .15, p (1) = .20, p (2) = .30, p (3) = .25, and p (4) = .10. Determine:
a) Is this a valid probability mass function? Why?
b) Obtain the cumulative probability function
c) What is the probability that at least two students come to counseling
d) What is the probability that, from one to three students, inclusive, they will come to counseling e) Obtain the average and standard deviation of the probability function. Interpret your results.
In: Statistics and Probability
15. A sample of 10 networking sites for a specific month has a mean number of visits of 26.1 and a standard deviation of 4.2. Find a 99 percent confidence interval of the true mean.
NOTE: be sure to take the square root of a given Variance in the word problems to obtain the standard deviation and use that for your figuring
a. 21.8, 30.4
b. 22.2, 28.4
c. 20.1, 33.2
d. 21.9, 36.8
In: Statistics and Probability
4. In a sample of 10 people in Japan average life expectancy is 85 years with a sample standard deviation of 3 years. In a sample of 8 people in South Korea, the corresponding numbers are 82 years and 2 years.
(a) Find a 95% confidence interval for the population average difference in life expectancy.
(b) Suppose you wish to test the null that the two population life expectancy values are equal against the alternative that they are different. Find a range in which the P-value must fall.
In: Statistics and Probability
In: Statistics and Probability
In 2011 home prices and mortgage rates dropped so low that in a number of cities the monthly cost of owning a home was less expensive than renting. The following data show the average asking rent for 10 markets and the monthly mortgage on the median priced home (including taxes and insurance) for 10 cities where the average monthly mortgage payment was less than the average asking rent (The Wall Street Journal, November 26–27, 2011). Click on the webfile logo to reference the data.
City | Rent ($) | Mortgage ($) | |
Atlanta | 840 | 539 | |
Chicago | 1062 | 1002 | |
Detroit | 823 | 626 | |
Jacksonville, Fla. | 779 | 711 | |
Las Vegas | 796 | 655 | |
Miami | 1071 | 977 | |
Minneapolis | 953 | 776 | |
Orlando, Fla. | 851 | 695 | |
Phoenix | 762 | 651 | |
St. Louis | 723 | 654 |
Enter negative values as negative numbers. a. Develop the estimated regression equation
that can be used to predict the monthly mortgage given the average
asking rent (to 2 decimals). b. Choose a residual plot against the
independent variable. SelectScatter diagram 1Scatter diagram 2Scatter diagram 3None of these choicesItem 3 c. Do the assumptions about the error term and
model form seem reasonable in light of the residual plot? |
Icon Key
Question 9 of 10
Exercise 12.49
In: Statistics and Probability
6. Suppose that the sample unemployment rate for those aged 25–60 is 8% based on a survey of 150 people, while the unemployment rate for those aged 16–24 is 12% based on a survey of 100 people.
(a) Form a 90% confidence interval for the difference between the two population unemployment rates.
(b) Construct a test statistic to test the null hypothesis that the two population rates are the same against the alternative hypothesis that the rate is higher in for those aged 16–24, and report the associated P-value.
In: Statistics and Probability
Two coins are tossed at the same time. Let random variable be the number of heads showing.
a) Construct a probability distribution for
b) Find the expected value of the number of heads.
In: Statistics and Probability
Student |
Critical Reading |
Mathematics |
Writing |
1 |
526 |
534 |
530 |
2 |
594 |
590 |
586 |
3 |
465 |
464 |
445 |
4 |
561 |
566 |
553 |
5 |
436 |
478 |
430 |
6 |
430 |
458 |
420 |
In: Statistics and Probability
Consider the following scenario and address the questions that follow.
Your leader has asked you to evaluate the defects based on insurance claim type (auto vs. injury). You have a total of 600 claims, with 400 auto, 200 injury, 100 with a defect, and 60 auto or has an injury.
In: Statistics and Probability
Assignment 3: Data Analysis with Graphs Assignment
1. how would you draw a stem and leaf diagram for each of the above data sets.
2.Using your stem and leaf diagrams, make at least 2 comments about each set of data (i.e., what does the diagram tell you about the data?).
3.For each of the data sets:
a.Create a histogram from the data.
b.Does the histogram tell you anything that wasn’t evident from the stem and leaf diagram? Does it make anything about the data more difficult to see?
c.Create a second histogram with a different bin width. How does this change how you interpret the data?
4.For each of the data sets, explain whether you prefer the stem and leaf diagram or a histogram and why?
In: Statistics and Probability