There are 3 SPSS outputs in this homework assignment. The questions for each output are listed below. Please type your answers into this word document and submit it as an attachment in the assignment tab.
Q1. Researchers were interested in determining whether background music helped or hindered students’ performance on a math test. Students were randomly assigned to 1 of 3 groups: 1) no music; 2) music only; and 3) music with lyrics. Students were then given a math exam, scores which could range from 0 to 100.
N |
Mean |
Std. Deviation |
Std. Error |
95% Confidence Interval for Mean |
||
Lower Bound |
Upper Bound |
|||||
No music |
250 |
77.59 |
13.055 |
.826 |
75.96 |
79.21 |
Music only |
250 |
78.10 |
13.357 |
.845 |
76.44 |
79.77 |
Music and lyrics |
250 |
78.97 |
13.263 |
.839 |
77.32 |
80.62 |
Total |
750 |
78.22 |
13.221 |
.483 |
77.27 |
79.17 |
ANOVA |
|||||
minutes |
|||||
Sum of Squares |
df |
Mean Square |
F |
Sig. |
|
Between Groups |
244.595 |
2 |
122.297 |
.699 |
.497 |
Within Groups |
130668.664 |
747 |
174.925 |
||
Total |
130913.259 |
749 |
In: Math
2. An experiment was conducted by a physiologist to determine whether exercise improves the human immune system. Thirty subjects volunteered to participate in the study. The amount of immunoglobulin known as IgG (an indicator of long-term immunity) and the maximal oxygen uptake (a measure of aerobic fitness level) were recorded for each subject. The data can be found in the file marked AEROBIC. You will need to use the Data Analysis - Regression Function for this problem, as well as some graphing functions.
a. Construct a scattergram for the IgG-maximal oxygen uptake data.
b. Hypothesize a probabilistic model relating IgG to maximal oxygen uptake.
c. Fit the model to the data. Is there sufficient evidence to indicate that the model provides information for the prediction of IgG, y? Test using α = .05.
d. Does a second-order term contribute information for the prediction of y? Test using α = .05.
Subject | IgG | Max Oxy |
1 | 881 | 34.6 |
2 | 1290 | 45.0 |
3 | 2147 | 62.3 |
4 | 1909 | 58.9 |
5 | 1282 | 42.5 |
6 | 1530 | 44.3 |
7 | 2067 | 67.9 |
8 | 1982 | 58.5 |
9 | 1019 | 35.6 |
10 | 1651 | 49.6 |
11 | 752 | 33.0 |
12 | 1687 | 52.0 |
13 | 1782 | 61.4 |
14 | 1529 | 50.2 |
15 | 969 | 34.1 |
16 | 1660 | 52.5 |
17 | 2121 | 69.9 |
18 | 1382 | 38.8 |
19 | 1714 | 50.6 |
20 | 1959 | 69.4 |
21 | 1158 | 37.4 |
22 | 965 | 35.1 |
23 | 1456 | 43.0 |
24 | 1273 | 44.1 |
25 | 1418 | 49.8 |
26 | 1743 | 54.4 |
27 | 1997 | 68.5 |
28 | 2177 | 69.5 |
29 | 1965 | 63.0 |
30 | 1264 | 43.2 |
In: Math
Use the table below to answer questions 4.5 – 4.7: This table contains the same client data as the first table. This time, though, the instructor is interested in knowing how his clients’ other activities might impact their average cycling speed in spin class. He notes that half of his clients also ride bikes outside during the week, while the other half of his clients do not bike anywhere except spin class.
Rides Outside |
Only Spin | Rides Outside | Only SPin | Rides outside | only spin |
---|---|---|---|---|---|
20 | 15 | ||||
17 | 17 | ||||
18 | 19 | ||||
22 | 17 | ||||
21 | 17 | ||||
18 | 16 | ||||
17 | 18 |
Average Speed M = 19 M = 17
4.5 Calculate SS for each sample of spin class clients (the portion who ride outside and the portion who only do spin class). Show Work by inserting numbers into the table to show intermediate steps Rides Outside SS = Only Spin Class SS =
4.6 Calculate s for each group Rides Outside s = Only Spin Class s =
4.7 Based on the statistics you have computed, does there appear to be any difference in average speed between those who bike outside and those who only bike during spin class?
Explain why or why not?
In: Math
Plot | Nutrients added | # of species |
1 | 0 | 36 |
2 | 0 | 36 |
3 | 0 | 32 |
4 | 1 | 34 |
5 | 2 | 33 |
6 | 3 | 30 |
7 | 1 | 20 |
8 | 3 | 23 |
9 | 4 | 21 |
10 | 4 | 16 |
What effect do nutrient additions have on plant species diversity? Long-term experiments at the Rothamstead Experimental Station in the U.K. sought to investigate the relationship, with some interesting findings.
The data can be found in the linked Google Sheets
document - you'll want to copy it to Excel and use the
Data Analysis ToolPak.
1) Produce a scatter plot of the data (click here for a generic
youtube video on creating a scatter plot from excel data - this is
for informational purposes only - it's not your data)
2) Add the least-squares regression line to your scatter plot. (click here for a generic youtube video on adding trendlines to scatter plots - this is for informational purposes only - it's not your data)
3) Test the hypothesis of no treatment effect on the number of plant species.
In: Math
A physician with a practice is currently serving 280 patients. The physician would like to administer a survey to his patients to measure their satisfaction level with his practice. A random sample of 22 patients had an average satisfaction score of 8.3 on a scale of 1-10. The sample standard deviation was 1.3 . Complete parts a and b below. a. Construct a 99% confidence interval to estimate the average satisfaction score for the physician's practice. The 99% confidence interval to estimate the average satisfaction score is left parenthesis nothing comma nothing right parenthesis . (Round to two decimal places as needed.)
In: Math
Recent incidents of food contamination have caused great concern among consumers. An article reported that 39 of 80 randomly selected Brand A brand chickens tested positively for either campylobacter or salmonella (or both), the leading bacterial causes of food-borne disease, whereas 62 of 80 Brand B brand chickens tested positive.
a)Does it appear that the true proportion of non-contaminated Brand A chickens differs from that for Brand B? Carry out a test of hypotheses using a significance level 0.01. (Use p1 for Brand A and p2 for Brand B.)
Calculate the test statistic and P-value. (Round your test statistic to two decimal places and your P-value to four decimal places.)
z= P-value =
b)if the true proportions of non-contaminated chickens for the Brand A and Brand B are 0.50 and 0.25, respectively, how likely is it that the null hypothesis of equal proportions will be rejected when a 0.01 significance level is used and the sample sizes are both 60? (Round your answer to four decimal places.)
In: Math
A supermarket is trying to decide whether to accept or reject a shipment of tomatoes. It is impossible to check all the tomatoes for size, but they desire an average weight of 8 ounces (they neither want too large nor too small).
(a) State the hypotheses.
(b) A random sample of 25 tomatoes yields an average weight of 7.65 ounces and a standard
deviation of 1.15 ounces. Calculate the test statistic and the p-value.
(c) Would you reject H0, or fail to reject H0 at 5% level of significance?
(d) Should the supermarket reject the shipment? Explain.
(e) To what type of error are you subject to?
In: Math
Collect data from 30 people from your work, school, neighborhood, family, or other group. Ask a quantitative question, such as, “How many pets do you have?” or “How many college classes have you taken?” Explain your population, sample, and sampling method and what level of measurement your data is (nominal, ordinal, interval, or ratio). Use technology ( Excel) to create a Histogram of your data and explain the shape of the distribution (bell-shaped, uniform, right-skewed, or left-skewed) and possible reasons why the distribution is this shape. Explain the importance of this data, what you find interesting about the data, and why the public should know. Look up a newspaper, e-pub, or journal article that confirms or denies the results of your small study.
Please explain briefly.
In: Math
7. A certain drug is used to treat asthma. In a clinical trial of the drug, 20 of 286 treated subjects experienced headaches (based on data from the manufacturer). The accompanying calculator display shows results from a test of the claim that less than 12% of treated subjects experienced headaches. Use the normal distribution as an approximation to the binomial distribution and assume a 0.05 significance level to complete parts (a) through (e) below.
a. Is the test two-tailed, left-tailed, or right-tailed?
Right tailed test
Left-tailed test
Two-tailed test
b. What is the test statistic?
Z=
(Round to two decimal places as needed.)
c. What is the P-value?
P-value=______
(Round to four decimal places as needed.)
d. What is the null hypothesis, and what do you conclude about it?
Identify the null hypothesis.
A:Ho<0.09
B:Ho>0.09
C:Ho:p≠0.09
D:Ho:p=0.09
Decide whether to reject the null hypothesis. Choose the correct answer below.
A.Reject the null hypothesis because the P-value is greater than the significance level, alpha.
B.Reject the null hypothesis because the P-value is less than or equal to the significance level, alpha.
C.Fail to reject the null hypothesis because the P-value is greater than the significance level, alpha.
D.Fail to reject the null hypothesis because the P-value is less than or equal to the significance level, alpha.
e. What is the final conclusion?
A.There is not sufficient evidence to support the claim that less than 9% of treated subjects experienced headaches.
B.There is not sufficient evidence to warrant rejection of the claim that less than 9% of treated subjects experienced headaches.
C.There is sufficient evidence to warrant rejection of the claim that less than 9% of treated subjects experienced headaches.
D.There is sufficient evidence to support the claim that less than 9% of treated subjects experienced headaches.
In: Math
Suppose μ1 and μ2 are true mean stopping distances at 50 mph for cars of a certain type equipped with two different types of braking systems. The data follows: m = 8,x = 115.8, s1 = 5.06, n = 8, y = 129.5, and s2 = 5.35. Calculate a 95% CI for the difference between true average stopping distances for cars equipped with system 1 and cars equipped with system 2. (Round your answers to two decimal places.)
In: Math
A machine that is programmed to package 3.90 pounds of cereal is being tested for its accuracy. In a sample of 64 cereal boxes, the sample mean filling weight is calculated as 3.95 pounds. The population standard deviation is known to be 0.14 pound. [You may find it useful to reference the z table.]
a-1. Identify the relevant parameter of interest for these quantitative data.
The parameter of interest is the average filling weight of all
cereal packages.
The parameter of interest is the proportion filling weight of all
cereal packages.
a-2. Compute its point estimate as well as the margin of error with 90% confidence. (Round intermediate calculations to at least 4 decimal places. Round "z" value to 3 decimal places and final answers to 2 decimal places.)
b-1. Calculate the 90% confidence interval. (Use rounded margin of error. Round your final answers to 2 decimal places.)
b-2. Can we conclude that the packaging machine is operating improperly?
No, since the confidence interval contains the target filling
weight of 3.90.
No, since the confidence interval does not contain the target
filling weight of 3.90.
Yes, since the confidence interval contains the target filling
weight of 3.90.
Yes, since the confidence interval does not contain the target
filling weight of 3.90.
c. How large a sample must we take if we want the margin of error to be at most 0.02 pound with 90% confidence? (Round intermediate calculations to at least 4 decimal places. Round "z" value to 3 decimal places and round up your final answer to the next whole number.)
In: Math
Problem 1
The dependent variable is assumed to be values of a land.
a) Use the Excel regression tool to do the linear regression, and provide the
“Line Fit Plots” (which is provided in the regression interface). (5 points)
b) What can the plot tell you? E.g., does it show that the fitting is good?
(5 points)
c) Now check the output.
c.1) What is the standard error of the estimate of the slope? (5 points)
c.2) What is the t-test statistic for the slope? Reproduce the t-test statistic
in Excel using other values in the output (e.g., point estimate and standard
error). (10 points)
c.2) What is the 95% confidence interval of the slope? Reproduce the con-
fidence interval in Excel using other values in the output (e.g., t stat). Can
we use the confidence interval to claim that the independent variable can be
dropped in the linear regression model? (10 points)
c.3) What is the p-value for the estimate of the slope? Reproduce the p-value
in Excel using other values in the output (e.g., t stat). Can we use the p-values
to claim that the independent variable can be dropped in the linear regression
model? (10 points)
c.4) Does the linear regression model fit well? Explain your answer. (5
points)
d) Assume that the area of the land you are considering to sell is only one
acre. Does the linear regression model provide a good prediction for the value
of your land? (5 points)
e) Assume that you want to check if the slope should be significantly bigger
than 10,000.
e.1) Write the hypotheses. (5 points)
e.2) What is the new t-test statistic? (5 points)
e.3) What is the new p-value for the estimate of the slope? Is the slope
significantly bigger than 10,000? (10 points)
Values |
836,586,000 |
986,547,000 |
1,075,609,000 |
381,443,000 |
889,148,000 |
1,096,422,000 |
1,340,628,000 |
903,129,000 |
785,261,000 |
1,407,381,000 |
799,722,000 |
1,242,590,000 |
378,638,000 |
395,110,000 |
582,299,000 |
286,805,000 |
1,286,312,000 |
188,313,000 |
529,053,000 |
700,357,000 |
1,123,597,000 |
392,277,000 |
1,068,679,000 |
576,348,000 |
535,527,000 |
797,064,000 |
854,322,000 |
1,415,763,000 |
1,110,576,000 |
543,485,000 |
621,503,000 |
44,632,000 |
473,953,000 |
129,286,000 |
372,399,000 |
604,300,000 |
432,818,000 |
748,532,000 |
139,826,000 |
456,433,000 |
1,694,543,000 |
967,926,000 |
1,009,765,000 |
1,085,302,000 |
1,089,378,000 |
1,331,657,000 |
364,124,000 |
1,070,730,000 |
1,536,796,000 |
1,426,503,000 |
796,188,000 |
1,559,685,000 |
493,466,000 |
743,640,000 |
376,926,000 |
957,234,000 |
169,340,000 |
157,625,000 |
309,507,000 |
265,410,000 |
251,621,000 |
412,789,000 |
136,533,000 |
184,032,000 |
256,578,000 |
228,716,000 |
565,330,000 |
219,363,000 |
388,716,000 |
81,059,000 |
371,794,000 |
853,684,000 |
618,448,000 |
1,032,717,000 |
876,501,000 |
157,428,000 |
726,993,000 |
1,178,550,000 |
762,332,000 |
1,269,773,000 |
1,018,473,000 |
895,709,000 |
2,412,768,000 |
1,211,090,000 |
1,060,153,000 |
2,145,334,000 |
1,050,692,000 |
1,227,843,000 |
In: Math
In: Math
5. The test statistic of z=2.31 is obtained when testing the claim that p>0.3.
a. This is a (two-tailed, right-tailed, left-tailed) test.
b. P-value=_____
(Round to three decimal places as needed.)
Choose the correct conclusion below.
A.Reject Ho. There is sufficient evidence to support the claim that p>0.3.
B.Reject Ho. There is not sufficient evidence to support the claim that p>0.3.
C.Fail to reject Ho. There is sufficient evidence to support the claim that p>0.3.
D.Fail to reject Ho. There is not sufficient evidence to support the claim that p>0.3
6. The test statistic of z=- 2.58 is obtained when testing the claim that P=3/5.
a. The critical value(s) is/are z=______
(Round to two decimal places as needed. Use a comma to separate answers as needed.)
b. Choose the correct conclusion below.
A.Reject Ho. There is not sufficient evidence to warrant rejection of the claim that P=3/5
B.Reject Ho. There is sufficient evidence to warrant rejection of the claim that P=3/5
C.Fail to reject Ho. There is sufficient evidence to warrant rejection of the claim that P=3/5
D.Fail to reject Ho. There is not sufficient evidence to warrant rejection of the claim that P=3/5
In: Math
Bass - Samples: The bass in Clear Lake have weights that are normally distributed with a mean of 2.2 pounds and a standard deviation of 0.6 pounds. Suppose you catch a stringer of 6 bass with a total weight of 15.9 pounds. Here we determine how unusual this is.
(a) What is the mean fish weight of your catch of 6? Round
your answer to 1 decimal place.
(b) If 6 bass are randomly selected from Clear Lake, find the
probability that the mean weight is greater than the mean of those
you caught. Round your answer to 4 decimal
places.
2
(c) Which statement best describes your situation?
This is unusual because the probability of randomly selecting 6 fish with a mean weight greater than or equal to the mean of your stringer is less than the benchmark probability of 0.05.
This is not particularly unusual because the mean weight of your fish is only 0.5 pounds above the population average.
In: Math