Louis katz, a cost accountant at papalote plastics, inc. (ppi), is
Louis katz, a cost accountant at papalote plastics, inc. (ppi), is
Question 13
Louis Katz, a cost accountant at Papalote Plastics, Inc. (PPI), is analyzing the manufacturing costs of a molded plastic telephone handset produced by PPI. Louis’s independent variable is production lot size (in 1,000’s of units), and his dependent variable is the total cost of the lot (in $100’s). Regression analysis of the data yielded the following tables.
|
|
Coefficients |
Standard Error |
t Statistic |
p-value |
|
||||||
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Intercept |
3.996 |
1.161268 |
3.441065 |
0.004885 |
|
||||||
|
x |
0.358 |
0.102397 |
3.496205 |
0.004413 |
|
||||||
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Source |
df |
SS |
MS |
F |
|
Se = 0.898 |
|||||
|
Regression |
1 |
9.858769 |
9.858769 |
12.22345 |
|
r2 = 0.526341 |
|||||
|
Residual |
11 |
8.872 |
0.806545 |
|
|
|
|||||
|
Total |
12 |
18.73077 |
|
|
|
|
|||||
Louis’s regression model is _____________.
|
|
y = -0.358 + 3.996x |
|
|
y = 3.996 – 0.358x |
|
|
y = 0.358 + 3.996x |
|
|
y = -3.996 + 0.358x |
|
|
y = 3.996 + 0.358x |
Question 20
According to a study several years ago by the Personal Communications Industry Association, the average wireless phone user earns $62,600 per year. Suppose a researcher believes that the average annual earnings of a wireless phone user are lower now, and he sets up a study in an attempt to prove his theory. He randomly samples 18 wireless phone users and finds out that the average annual salary for this sample is $58,974, with a population standard deviation of $7,810. Use α = .01to test the researcher’s theory. Assume wages in this industry are normally distributed.
The value of the test statistic rounded to 2 decimal places is z =
and we
.
The tolerance is +/- 0.05.
A random sample of 51 items is taken, with and s2 = 25.68. Use these data to test the following hypotheses, assuming you want to take only a 1% risk of committing a Type I error and that x is normally distributed.
StartLayout 1st Row 1st Column Upper H Subscript 0 Baseline colon mu equals 60 EndLayout StartLayout 1st Row 1st Column Upper H Subscript a Baseline colon mu less-than 60 EndLayout
Round your answer to 2 decimal places, the tolerance is +/-0.01.
The value of the test statistic is
and we
.
Question 34
Louis Katz, a cost accountant at Papalote Plastics, Inc. (PPI), is analyzing the manufacturing costs of a molded plastic telephone handset produced by PPI. Louis’s independent variable is production lot size (in 1,000’s of units), and his dependent variable is the total cost of the lot (in $100’s). Regression analysis of the data yielded the following tables.
|
|
Coefficients |
Standard Error |
t Statistic |
p-value |
|
||||||
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Intercept |
3.996 |
1.161268 |
3.441065 |
0.004885 |
|
||||||
|
x |
0.358 |
0.102397 |
3.496205 |
0.004413 |
|
||||||
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Source |
df |
SS |
MS |
F |
|
Se = 0.898 |
|||||
|
Regression |
1 |
9.858769 |
9.858769 |
12.22345 |
|
r2 = 0.526341 |
|||||
|
Residual |
11 |
8.872 |
0.806545 |
|
|
|
|||||
|
Total |
12 |
18.73077 |
|
|
|
|
|||||
Using α = 0.05, Louis should ___________.
|
|
not reject H0: β1 = 0 |
|
|
reject H0: β1 = 0 |
|
|
do not reject H0: β0 = 0 |
|
|
suspend judgment |
|
|
increase the sample size |
Louis katz, a cost accountant at papalote plastics, inc. (ppi), is
Question 13
Louis Katz, a cost accountant at Papalote Plastics, Inc. (PPI), is analyzing the manufacturing costs of a molded plastic telephone handset produced by PPI. Louis’s independent variable is production lot size (in 1,000’s of units), and his dependent variable is the total cost of the lot (in $100’s). Regression analysis of the data yielded the following tables.
|
|
Coefficients |
Standard Error |
t Statistic |
p-value |
|
||||||
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Intercept |
3.996 |
1.161268 |
3.441065 |
0.004885 |
|
||||||
|
x |
0.358 |
0.102397 |
3.496205 |
0.004413 |
|
||||||
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Source |
df |
SS |
MS |
F |
|
Se = 0.898 |
|||||
|
Regression |
1 |
9.858769 |
9.858769 |
12.22345 |
|
r2 = 0.526341 |
|||||
|
Residual |
11 |
8.872 |
0.806545 |
|
|
|
|||||
|
Total |
12 |
18.73077 |
|
|
|
|
|||||
Louis’s regression model is _____________.
|
|
y = -0.358 + 3.996x |
|
|
y = 3.996 – 0.358x |
|
|
y = 0.358 + 3.996x |
|
|
y = -3.996 + 0.358x |
|
|
y = 3.996 + 0.358x |
Question 20
According to a study several years ago by the Personal Communications Industry Association, the average wireless phone user earns $62,600 per year. Suppose a researcher believes that the average annual earnings of a wireless phone user are lower now, and he sets up a study in an attempt to prove his theory. He randomly samples 18 wireless phone users and finds out that the average annual salary for this sample is $58,974, with a population standard deviation of $7,810. Use α = .01to test the researcher’s theory. Assume wages in this industry are normally distributed.
The value of the test statistic rounded to 2 decimal places is z =
and we
.
The tolerance is +/- 0.05.
A random sample of 51 items is taken, with and s2 = 25.68. Use these data to test the following hypotheses, assuming you want to take only a 1% risk of committing a Type I error and that x is normally distributed.
StartLayout 1st Row 1st Column Upper H Subscript 0 Baseline colon mu equals 60 EndLayout StartLayout 1st Row 1st Column Upper H Subscript a Baseline colon mu less-than 60 EndLayout
Round your answer to 2 decimal places, the tolerance is +/-0.01.
The value of the test statistic is
and we
.
Question 34
Louis Katz, a cost accountant at Papalote Plastics, Inc. (PPI), is analyzing the manufacturing costs of a molded plastic telephone handset produced by PPI. Louis’s independent variable is production lot size (in 1,000’s of units), and his dependent variable is the total cost of the lot (in $100’s). Regression analysis of the data yielded the following tables.
|
|
Coefficients |
Standard Error |
t Statistic |
p-value |
|
||||||
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Intercept |
3.996 |
1.161268 |
3.441065 |
0.004885 |
|
||||||
|
x |
0.358 |
0.102397 |
3.496205 |
0.004413 |
|
||||||
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Source |
df |
SS |
MS |
F |
|
Se = 0.898 |
|||||
|
Regression |
1 |
9.858769 |
9.858769 |
12.22345 |
|
r2 = 0.526341 |
|||||
|
Residual |
11 |
8.872 |
0.806545 |
|
|
|
|||||
|
Total |
12 |
18.73077 |
|
|
|
|
|||||
Using α = 0.05, Louis should ___________.
|
|
not reject H0: β1 = 0 |
|
|
reject H0: β1 = 0 |
|
|
do not reject H0: β0 = 0 |
|
|
suspend judgment |
|
|
increase the sample size |