Statisctics
Running Head: STATISTICS 1
STATISTICS 11
STATISTICS
NAME
PROFESSOR NAME
DATE DUE
Activity 1
Ball bearing manufacturing is a highly precise business in which minimal part variability is critical. Large variances in the size of the ball bearings cause bearing failure and rapid wear-out. Production standards call for a maximum variance of .0001 inches. Gerry Liddy has gathered a sample of 15 bearings that show a sample standard deviation of.014inches. use a=.10
Please determine whether the sample indicates that the maximum acceptable variance is being exceeded.
Yes, the maximum acceptable variance is being exceeded.
What is the p value?
P =0.000196
2. the grade point averages of 352 students who completed a collage course in financial accounting have a standard deviation of .940 the grade point averages of 73 students who dropped out of the same course have a standard deviation of .797
a. Do the data indicate a difference between the variance of grade point averaged for student who completed a financial accounting course and students who dropped out?
Yes, it does.
It indicates a difference where the variance of students who completed collage is 0.8836 while the variance of those who dropped out is 0.6352. which gives a difference of 0.2484.
use a=.05 level of significance, what is the p value?
0.940+0.797=1.737
352 students
Std = 0.940
73 drop out students
Std = 0.797
P value 0.940-0.797 = 0.143
Activity 2
Use the sample data provided in table 1 to answer the following questions:
Calculate the point estimate of the population proportion of visitors who rated each of these museums as spectacular.
88
Conduct a hypothesis test to determine if the population of visitors who rated the museum as spectacular is equal for these five museums using a=.05 level of significance.
What is the p value?
H0= population that rated the museums spectacular is >88
H1= population that rated the museum spectacular + not spectacular = 1
88+44=132
132=1
88=?
0.6667
If the null is rejected perform post-hoc test using a and make conclusions.
The null is not rejected.
Table 2
Use the sample data provided in the table 2 to answer the following questions
How large was the sample in thus poll?
1087+1076+1045+1109+1120+1020= 6457
=6457
Conduct a hypothesis test to determine whether people’s attitude toward building new nuclear power plants is independent of country using a= .05 level of significance. What is the p value and what is your conclusion?
P of strongly favor = 141+161+298+133+128+204 = 1065 / total population
1065/6457
0.1649
Using the percentage of respondents who strongly favor and favor more than oppose which country has the most favorable attitude toward building new nuclear power plants? Which country has the least favorable attitude?
Most favorable – United States
204+326=530
(530/1020) *100 = 51.96%
Least favorable – Spain
133+222=355
355/1109*100 = 32.01%
A sample of 400 compact car sales in Chicago showed the number of vehicles sold in table 3
Table 3
|
Honda civic |
98 |
|
Toyota corolla |
72 |
|
Nissan sentra |
54 |
|
Hyundai Elantra |
44 |
|
Chevrolet Cruze |
42 |
|
Ford focus |
25 |
|
others |
65 |
Use a goodness of fit test to determine if the sample data indicate that the market shares for compact cars in Chicago are different than the market shares suggested by nationwide 2017 sales using a=.05 level of significance.
What is the p value and what is your conclusion?
P >25
There has been a great increase in the car sales.
If the Chicago market appears to differ significantly from the nationwide sales, which categories contribute most to this difference? -OTHERS
Activity 3
|
Tire |
steering |
Treadwear |
Buy again |
|
Goodyear assurance triplet red |
8.9 |
8.5 |
8.1 |
|
Michelin hydro edge |
8.9 |
9.0 |
8.3 |
|
Michelin harmony |
8.3 |
8.8 |
8.2 |
|
Dunlop SP 60 |
8.2 |
8.5 |
7.9 |
|
Goodyear assurance comforted |
7.9 |
7.7 |
7.1 |
|
Yokohama Y372 |
8.4 |
8.2 |
8.9 |
|
Yokohama aegis LS4 |
7.9 |
7.0 |
7.1 |
|
Kumho power star |
7.9 |
7.9 |
8.3 |
|
Goodyear assurance |
7.6 |
5.8 |
4.5 |
|
Hankook H406 |
7.8 |
6.8 |
6.2 |
|
Michelin Energy LX4 |
7.4 |
5.7 |
4.8 |
|
Michelin MX4 |
7.0 |
6.5 |
5.3 |
|
Micheline symmetric |
6.9 |
5.7 |
4.2 |
|
Dunlop SP 40 A/S |
6.2 |
4.2 |
3.4 |
|
Bridgestone insignia SE20 |
5.7 |
5.5 |
3.6 |
|
Goodyear integrity |
5.7 |
5.4 |
2.9 |
|
Dunlop SP20FE |
5.7 |
5.5 |
3.3 |
|
Kamuho 772 |
7.2 |
6.6 |
5.0 |
Use the data provided in table 4
Provide descriptive statistics of the data.
Mean of the steerling = 7.4
Mean of the treadwear = 6.9
Mean of buy again = 5.8
Median of steerling = 7.8
Median of treadwear = 6.3
Median of buy again =5.4
Develop two simple regression models that can be used to predict the buy again rating given the steering rating in one and the tread wear rating in the other. State the hypotheses on the coefficients, justify formulations of these hypotheses, and interpret the result. Use a=.05. include all phases of assessment of the model.
|
Tire |
steering |
Treadwear |
Buy again |
|
Yokohama Y372 |
8.4 |
8.2 |
8.9 |
|
Michelin harmony |
8.3 |
8.8 |
8.2 |
H0= buy again 8.9
H1=buy again 8.2
8.9 + 8.2= 17.1
17.1/123.1
=0.1389
Develop a multiple regression model that can be used to predict the buy again rating given the steering rating and the tread wear rating. State the hypotheses on the coefficients, justify formulations of these hypotheses, and interept the result. Use a = .05 include all phases of assessment of the model and do not forget to check multicollinearity.
|
Tire |
steering |
Treadwear |
Buy again |
|
Goodyear assurance triplet red |
8.9 |
8.5 |
8.1 |
|
Michelin hydro edge |
8.9 |
9.0 |
8.3 |
|
Michelin harmony |
8.3 |
8.8 |
8.2 |
|
Dunlop SP 60 |
8.2 |
8.5 |
7.9 |
|
Goodyear assurance comforted |
7.9 |
7.7 |
7.1 |
8.1+8.3+8.2+7.9+7.1=39.6
39.6/123.1 =0.3217
Does combining the two independent variables improve coefficients of determination? Please explain.
Yes, it improves coefficients.
Adding more terms inherently improves the fit. It gives a new term for the model to use to fit the data and a new coefficient that it can vary to force a better fit.
Choose a combination of steering and tread wear not given in the above table and find the expected buy again for this combination.
The buy again could be very low compared to when the steering and tread wear given because people don’t see the qualities of the tires.
Professional assignment
Provide descriptive statistics of the data, develop a multiple regression model that can be used to predict the amount charged given the annual income and household size, state the hypotheses on the coefficients, justify formulation of these hypotheses, and interpret the results. Use a=.05. include all phases of assessment of the model and do not forget to check multicollinearity.
|
Income ($1000 |
Household size |
Amount charged($) |
Income |
Size |
Charged |
|
54 |
3 |
4016 |
54 |
6 |
5573 |
|
30 |
2 |
3159 |
30 |
1 |
2583 |
|
32 |
4 |
5100 |
48 |
2 |
3866 |
|
50 |
5 |
4742 |
34 |
5 |
3586 |
H0= the higher the income the higher the amount charged
H1= the income is proportional to the amount charged
To assess robustness of the software, repeat part a but this time use full representation of annual income. What conclusions can you make?
|
Income ($1000 |
Household size |
Amount charged($) |
Income |
Size |
Charged |
|
54 |
3 |
4016 |
54 |
6 |
5573 |
|
30 |
2 |
3159 |
30 |
1 |
2583 |
|
32 |
4 |
5100 |
48 |
2 |
3866 |
|
50 |
5 |
4742 |
34 |
5 |
3586 |
We conclude that as the income increases the amount charged also increases.
Choose a combination of annual income and household size not given in the table and find the expected amount charged for this combination. Use a = .05 in testing all hypotheses
|
Income ($1000) |
Household size |
Amount charged($) |
Income |
Size |
Charged |
|
30 |
3 |
4056 |
30 |
6 |
5576 |
|
54 |
2 |
3044 |
54 |
1 |
3465 |
|
40 |
4 |
6503 |
40 |
2 |
2583 |
|
60 |
2 |
9433 |
60 |
3 |
3467 |