Please note that for all problems in this course, the standard cut-off (alpha) for a test of significance will be .05, and you always report the exact power unless SPSS output states p=.000 (you’d report p<.001). Also, remember when hand-calculating, always use TWO decimal places so that deductions in grading won’t be due to rounding differences.
Problem Set 1: (22 pts) A teacher wanted to see if a new pedagogical approach was beneficial to students, and if the effects vary by amount math anxiety. To examine this, students completed math anxiety surveys and were grouped into “low”, “average”, and “high” levels of math anxiety. She then randomly assigned equal numbers of students from each group into one of two classes – one taught using her “standard” approach; the other class was taught using a new approach. The data presented in the table below are the final grade percentages of the 60 children (30 per class). Conduct the most appropriate statistical analysis to determine whether final grades in a math class were affected by teaching approach and/or amount of math anxiety.
|
In: Math
Do heavier cars really use more gasoline? Suppose a car is
chosen at random. Let x be the weight of the car (in hundreds of
pounds), and let y be the miles per gallon (mpg).
x 27 42 31 47 23 40 34 52
y 33 21 24 13 29 17 21 14
Complete parts (a) through (e), given Σx = 296, Σy = 172, Σx2 = 11,652, Σy2 = 4042, Σxy = 5917, and r ≈ −0.911. (a) Draw a scatter diagram displaying the data.
(b) Verify the given sums Σx, Σy, Σx2, Σy2, Σxy, and the value of the sample correlation coefficient r. (Round your value for r to three decimal places.)
Σx =2 Σy =3 Σx2 =4 Σy2 =5 Σxy =6 r =7
(c) Find x, and y. Then find the equation of the least-squares line = a + bx. (Round your answers for x and y to two decimal places. Round your answers for a and b to three decimal places.) x = ___ y = ___ = y____ + ____x
(d) Graph the least-squares line. Be sure to plot the point (x, y) as a point on the line.
(e) Find the value of the coefficient of determination r2. What percentage of the variation in y can be explained by the corresponding variation in x and the least-squares line? What percentage is unexplained? (Round your answer for r2 to three decimal places. Round your answers for the percentages to one decimal place.)
r2 = _______ explained _____ % unexplained ______ %
(f) Suppose a car weighs x = 39 (hundred pounds). What does the least-squares line forecast for y = miles per gallon? (Round your answer to two decimal places.) ________mpg
In: Math
Assess this statement using ANOVA: "People with different levels of education exercise for different amounts of time during the week."
Describe which ANOVA test you used and present the main results, including explaining the within and between subjects variation and the F-ratio from the ANOVA table. Discuss whether there is a statistically significant difference between education groups and the amount of exercise.
Education_Level*** |
2 |
2 |
3 |
1 |
1 |
2 |
3 |
1 |
2 |
2 |
3 |
1 |
1 |
3 |
2 |
1 |
2 |
3 |
2 |
1 |
2 |
3 |
2 |
1 |
1 |
3 |
3 |
1 |
3 |
3 |
Minutes_Exercise |
90 |
50 |
65 |
20 |
50 |
25 |
110 |
50 |
40 |
80 |
120 |
80 |
60 |
150 |
75 |
80 |
110 |
80 |
100 |
0 |
50 |
200 |
60 |
65 |
40 |
65 |
70 |
45 |
75 |
50 |
In: Math
The following data was collected to explore how the average number of hours a student studies per night and the student's GPA affect their ACT score. The dependent variable is the ACT score, the first independent variable (x1) is the number of hours spent studying, and the second independent variable (x2) is the student's GPA.
Study Hours | GPA | ACT Score |
---|---|---|
0 | 2 | 16 |
1 | 2 | 19 |
3 | 2 | 24 |
4 | 3 | 27 |
5 | 3 | 30 |
Find the p-value for the regression equation that fits the given data. Round your answer to four decimal places.
In: Math
Round Tree Manor is a hotel that provides two types of rooms with three rental classes: Super Saver, Deluxe, and Business. The profit per night for each type of room and rental class is as follows:
Rental Class |
||||
|
Super Saver |
Deluxe |
Business |
|
Type I |
$38 |
$43 |
— |
|
Type II |
$16 |
$28 |
$45 |
Type I rooms do not have wireless Internet access and are not available for the Business rental class.
Round Tree's management makes a forecast of the demand by rental class for each night in the future. A linear programming model developed to maximize profit is used to determine how many reservations to accept for each rental class. The demand forecast for a particular night is 140 rentals in the Super Saver class, 50 rentals in the Deluxe class, and 40 rentals in the Business class. Round Tree has 110 Type I rooms and 110 Type II rooms.
Variable |
# of reservations |
SuperSaver rentals allocated to room type I |
|
SuperSaver rentals allocated to room type II |
|
Deluxe rentals allocated to room type I |
|
Deluxe rentals allocated to room type II |
|
Business rentals allocated to room type II |
Rental Class |
# of reservations |
SuperSaver |
|
Deluxe |
|
Business |
In: Math
You have brown, pink, black, and white socks in a drawer (8 of each color). In each of the following cases, what is the minimum number that you must take out to ensure that:
a) You have a matching pair?
b) You have two of different colors?
c) You have at least 3 brown or 4 pink or 5 black or 7 white socks?
In: Math
Problem 2
A random sample of 41 business days, from 2016 through 2017, of the closing price of Apple stock is conducted. The sample produces an average closing price of $116.16 with a standard deviation of $10.27.
In: Math
The following data represent a sample of the assets (in millions of dollars) of 28 credit unions in a state. Assume that the population in this state is normally distributed with σ=3.5 million dollars. Use Excel to find the 99% confidence interval of the mean assets in millions of dollars. Round your answers to three decimal places and use ascending order.
Assets
12.23
2.89
13.19
73.25
11.59
8.74
7.92
40.22
5.01
2.27
16.56
1.24
9.16
1.91
6.69
3.17
4.78
2.42
1.47
12.77
4.39
2.17
1.42
14.64
1.06
18.13
16.85
21.58
Provide your answer below: ( , )
In: Math
The following data was collected to explore how the number of square feet in a house, the number of bedrooms, and the age of the house affect the selling price of the house. The dependent variable is the selling price of the house, the first independent variable (x1x1) is the square footage, the second independent variable (x2x2) is the number of bedrooms, and the third independent variable (x3x3) is the age of the house.
Square Feet | Number of Bedrooms | Age | Selling Price |
---|---|---|---|
2032 | 4 | 9 | 262300 |
1101 | 3 | 8 | 182200 |
1526 | 5 | 9 | 176700 |
1612 | 3 | 15 | 265500 |
2607 | 2 | 2 | 257100 |
2077 | 3 | 10 | 255000 |
1277 | 3 | 13 | 275600 |
2048 | 3 | 10 | 156800 |
2339 | 3 | 9 | 290200 |
Copy Data
Find the p-value for the regression equation that fits the given data. Round your answer to four decimal places.
If the relationship is statistically significant, indicate the multiple regression equation that best fits the data, rounded to 3 decimal places.
In: Math
The following table shows ceremonial ranking and type of pottery sherd for a random sample of 434 sherds at an archaeological location.
Ceremonial Ranking | Cooking Jar Sherds | Decorated Jar Sherds (Noncooking) | Row Total |
A | 90 | 45 | 135 |
B | 96 | 49 | 145 |
C | 80 | 74 | 154 |
Column Total | 266 | 168 | 434 |
Use a chi-square test to determine if ceremonial ranking and pottery type are independent at the 0.05 level of significance.
(a) What is the level of significance?
State the null and alternate hypotheses.
H0: Ceremonial ranking and pottery type are
independent.
H1: Ceremonial ranking and pottery type are not
independent. H0: Ceremonial ranking and pottery
type are independent.
H1: Ceremonial ranking and pottery type are
independent. H0:
Ceremonial ranking and pottery type are not independent.
H1: Ceremonial ranking and pottery type are
independent. H0: Ceremonial ranking and pottery
type are not independent.
H1: Ceremonial ranking and pottery type are not
independent.
(b) Find the value of the chi-square statistic for the sample.
(Round the expected frequencies to at least three decimal places.
Round the test statistic to three decimal places.)
Are all the expected frequencies greater than 5?
Yes No
What sampling distribution will you use?
Student's t chi-square uniform binomial normal
What are the degrees of freedom?
(c) Find or estimate the P-value of the sample test
statistic. (Round your answer to three decimal places.)
p-value > 0.100 0.050 < p-value < 0.100 0.025 < p-value < 0.050 0.010 < p-value < 0.025 0.005 < p-value < 0.010 p-value < 0.005
(d) Based on your answers in parts (a) to (c), will you reject or
fail to reject the null hypothesis of independence?
Since the P-value > α, we fail to reject the null hypothesis. Since the P-value > α, we reject the null hypothesis. Since the P-value ≤ α, we reject the null hypothesis. Since the P-value ≤ α, we fail to reject the null hypothesis.
(e) Interpret your conclusion in the context of the
application.
At the 5% level of significance, there is sufficient evidence to conclude that ceremonial ranking and pottery type are not independent. At the 5% level of significance, there is insufficient evidence to conclude that ceremonial ranking and pottery type are not independent.
In: Math
Two plots at Rothamsted Experimental Station were studied for production of wheat straw. For a random sample of years, the annual wheat straw production (in pounds) from one plot was as follows.
6.19 | 5.98 | 5.77 | 7.10 | 7.31 | 7.18 |
7.06 | 5.79 | 6.24 | 5.91 | 6.14 |
Use a calculator to verify that, for this plot, the sample
variance is s2 ≈ 0.368.
Another random sample of years for a second plot gave the following
annual wheat production (in pounds).
5.91 | 5.77 | 6.12 | 6.96 | 7.22 | 5.58 | 5.47 | 5.86 |
Use a calculator to verify that the sample variance for this
plot is s2 ≈ 0.409.
Test the claim that there is a difference (either way) in the
population variance of wheat straw production for these two plots.
Use a 5% level of signifcance.
(a) What is the level of significance?
State the null and alternate hypotheses.
Ho: σ12 = σ22; H1: σ12 > σ22Ho: σ12 > σ22; H1: σ12 = σ22 Ho: σ22 = σ12; H1: σ22 > σ12Ho: σ12 = σ22; H1: σ12 ≠ σ22
(b) Find the value of the sample F statistic. (Use 2
decimal places.)
What are the degrees of freedom?
dfN | |
dfD |
What assumptions are you making about the original distribution?
The populations follow independent normal distributions. The populations follow independent normal distributions. We have random samples from each population. The populations follow dependent normal distributions. We have random samples from each population. The populations follow independent chi-square distributions. We have random samples from each population.
(c) Find or estimate the P-value of the sample test
statistic. (Use 4 decimal places.)
p-value > 0.200 0.100 < p-value < 0.200 0.050 < p-value < 0.100 0.020 < p-value < 0.050 0.002 < p-value < 0.020 p-value < 0.002
(d) Based on your answers in parts (a) to (c), will you reject or
fail to reject the null hypothesis?
At the α = 0.05 level, we reject the null hypothesis and conclude the data are not statistically significant. At the α = 0.05 level, we reject the null hypothesis and conclude the data are statistically significant. At the α = 0.05 level, we fail to reject the null hypothesis and conclude the data are not statistically significant. At the α = 0.05 level, we fail to reject the null hypothesis and conclude the data are statistically significant.
(e) Interpret your conclusion in the context of the
application.
Fail to reject the null hypothesis, there is sufficient evidence that the variance in annual wheat production differs between the two plots. Reject the null hypothesis, there is insufficient evidence that the variance in annual wheat production differs between the two plots. Reject the null hypothesis, there is sufficient evidence that the variance in annual wheat production differs between the two plots. Fail to reject the null hypothesis, there is insufficient evidence that the variance in annual wheat production differs between the two plots.
In: Math
The age distribution of the Canadian population and the age distribution of a random sample of 455 residents in the Indian community of a village are shown below.
Age (years) | Percent of Canadian Population | Observed Number in the Village |
Under 5 | 7.2% | 50 |
5 to 14 | 13.6% | 81 |
15 to 64 | 67.1% | 284 |
65 and older | 12.1% | 40 |
Use a 5% level of significance to test the claim that the age distribution of the general Canadian population fits the age distribution of the residents of Red Lake Village.
(a) What is the level of significance?
State the null and alternate hypotheses.
H0: The distributions are different.
H1: The distributions are different.
H0: The distributions are different.
H1: The distributions are the
same. H0: The
distributions are the same.
H1: The distributions are different.
H0: The distributions are the same.
H1: The distributions are the same.
(b) Find the value of the chi-square statistic for the sample.
(Round your answer to three decimal places.)
Are all the expected frequencies greater than 5?
Yes No
What sampling distribution will you use?
uniform chi-square binomial normal Student's t
What are the degrees of freedom?
(c) Estimate the P-value of the sample test statistic.
P-value > 0.100 0.050 < P-value < 0.100 0.025 < P-value < 0.050 0.010 < P-value < 0.025 0.005 < P-value < 0.010 P-value < 0.005
(d) Based on your answers in parts (a) to (c), will you reject or
fail to reject the null hypothesis that the population fits the
specified distribution of categories?
Since the P-value > α, we fail to reject the null hypothesis. Since the P-value > α, we reject the null hypothesis. Since the P-value ≤ α, we reject the null hypothesis. Since the P-value ≤ α, we fail to reject the null hypothesis.
(e) Interpret your conclusion in the context of the
application.
At the 5% level of significance, the evidence is insufficient to conclude that the village population does not fit the general Canadian population. At the 5% level of significance, the evidence is sufficient to conclude that the village population does not fit the general Canadian population.
In: Math
In: Math
How does gender and occupational prestige affect credibility?
Graduate students in a public health program are asked to rate the
strength of a paper about the health risks of childhood obesity. In
reality, all student raters are given the same paper, but the name
and degree associated with the author are changed. The student
raters are randomly assigned to one group from the following name
("John Lake", "Joan Lake") and degree (M.D., R.N., Ph.D.)
combination. The raters score the paper from 1 to 5 on clarity,
strength of argument, and thoroughness. The total scores (the sum
of the three scores) are given in the table below. What can be
concluded with an α of 0.05?
John Lake / M.D. |
John Lake / R.N. |
John Lake / Ph.D. |
Joan Lake / M.D. |
Joan Lake / R.N. |
Joan Lake / Ph.D. |
---|---|---|---|---|---|
12 15 13 15 14 |
15 16 12 13 13 |
11 8 13 12 9 |
15 10 12 14 12 |
10 7 8 10 6 |
11 11 12 8 8 |
Compute the corresponding effect size(s) and indicate
magnitude(s).
Name: η2
= ; ---Select--- na trivial effect
small effect medium effect large effect
Degree: η2
= ; ---Select--- na trivial effect
small effect medium effect large effect
Interaction: η2
= ; ---Select--- na trivial effect
small effect medium effect large effect
In: Math
High-profile legal cases have many people reevaluating the jury system. Many believe that juries in criminal trials should be able to convict on less than a unanimous vote. To assess support for this idea, investigators asked each individual in a random sample of Californians whether they favored allowing conviction by a 10–2 verdict in criminal cases not involving the death penalty. A newspaper article reported that 72% supported the 10–2 verdict. Suppose that the sample size for this survey was n = 900.Compute a 99% confidence interval for the proportion of Californians who favor the 10–2 verdict.
(________,________) |
Interpret the interval.
We are confident that 99% of the proportion of Californians who favor the 10–2 verdict is within this interval.
We are 99% confident that the proportion of all people who favor the 10–2 verdict is within this interval.
We are confident that the proportion of all people who favor the 10–2 verdict is within this interval at least 99% of the time.
We are 99% confident that the proportion of Californians who favor the 10–2 verdict is within this interval.
In: Math