The Nero Match Company sells matchboxes that are said to contain an average of 40 matches per box with a standard deviation of 9 matches. A random sample of 94 Nero matchboxes shows the average number per box to be 41.8 Using a 1% level of significance, can you say that the average number of matches per box is more than 40?
1) What type of hypothesis test is this?
2) State the hypotheses.
3) Calculate the value of the Test Statistic.
4) Calculate the P-Value.
5) What is the decision?
In: Statistics and Probability
The Nero Match Company sells matchboxes that are supposed to
have an average of 40 matches per box, with σ = 8. A
random sample of 92 matchboxes shows the average number of matches
per box to be 42.9. Using a 1% level of significance, can you say
that the average number of matches per box is more than 40?
What are we testing in this problem?
single proportion
single mean
(a) What is the level of significance?
State the null and alternate hypotheses.
H0: μ = 40; H1: μ ≠ 40
H0: p = 40; H1: p < 40
H0: μ = 40; H1: μ < 40
H0: μ = 40; H1: μ > 40
H0: p = 40; H1: p > 40
H0: p = 40; H1: p ≠ 40
(b) What sampling distribution will you use? What assumptions are
you making?
The Student's t, since we assume that x has a normal distribution with known σ.
The standard normal, since we assume that x has a normal distribution with known σ.
The standard normal, since we assume that x has a normal distribution with unknown σ.
The Student's t, since we assume that x has a normal distribution with unknown σ.
What is the value of the sample test statistic? (Round your answer
to two decimal places.)
(c) Find (or estimate) the P-value.
P-value > 0.250
0.125 < P-value < 0.250
0.050 < P-value < 0.125
0.025 < P-value < 0.050
0.005 < P-value < 0.025
P-value < 0.005
Sketch the sampling distribution and show the area corresponding to
the P-value.
(d) Based on your answers in parts (a) to (c), will you reject or
fail to reject the null hypothesis? Are the data statistically
significant at level α?
At the α = 0.01 level, we reject the null hypothesis and conclude the data are statistically significant.
At the α = 0.01 level, we reject the null hypothesis and conclude the data are not statistically significant.
At the α = 0.01 level, we fail to reject the null hypothesis and conclude the data are statistically significant.
At the α = 0.01 level, we fail to reject the null hypothesis and conclude the data are not statistically significant.
(e) Interpret your conclusion in the context of the
application.
There is sufficient evidence at the 0.01 level to conclude that the average number of matches per box is now greater than 40.
There is insufficient evidence at the 0.01 level to conclude that the average number of matches per box is now greater than 40.
In: Statistics and Probability
The Nero Match Company sells matchboxes that are supposed to
have an average of 40 matches per box, with σ = 8. A
random sample of 90 matchboxes shows the average number of matches
per box to be 42.6. Using a 1% level of significance, can you say
that the average number of matches per box is more than 40?
a. What are we testing in this problem?
single mean
single proportion
b. What is the level of significance?
c. State the null and alternate hypotheses.
H0: μ ≥ 40; H1: μ < 40
H0: μ = 40; H1: μ ≠ 40
H0: p = 40; H1: p ≠ 40
H0: p ≤ 40; H1: p > 40
H0: μ ≤ 40; H1: μ > 40
H0: p ≥ 40; H1: p < 40
d. What sampling distribution will you use? What
assumptions are you making?
The standard normal, since we assume that x has a normal distribution with known σ.
The standard normal, since we assume that x has a normal distribution with unknown σ.
The Student's t, since we assume that x has a normal distribution with known σ.
The Student's t, since we assume that x has a normal distribution with unknown σ.
e. What is the value of the sample test statistic? (Round
your answer to two decimal places.)
f. Estimate the P-value.
P-value > 0.250
0.125 < P-value < 0.250
0.050 < P-value < 0.125
0.025 < P-value < 0.050
0.005 < P-value < 0.025
P-value < 0.005
g. Sketch the sampling distribution and show the area
corresponding to the P-value.
h. Will you reject or fail to reject the null hypothesis?
Are the data statistically significant at level
α?
At the α = 0.01 level, we reject the null hypothesis and conclude the data are statistically significant.
At the α = 0.01 level, we reject the null hypothesis and conclude the data are not statistically significant.
At the α = 0.01 level, we fail to reject the null hypothesis and conclude the data are statistically significant.
At the α = 0.01 level, we fail to reject the null hypothesis and conclude the data are not statistically significant.
i. Interpret your conclusion in the context of the
application.
There is sufficient evidence at the 0.01 level to conclude that the average number of matches per box is now greater than 40.
There is insufficient evidence at the 0.01 level to conclude that the average number of matches per box is now greater than 40.
In: Statistics and Probability
The Nero Match Company sells matchboxes that are supposed to have an average of 40 matches per box, with σ = 8. A random sample of 90 matchboxes shows the average number of matches per box to be 43.0. Using a 1% level of significance, can you say that the average number of matches per box is more than 40? What are we testing in this problem? single proportion single mean (a) What is the level of significance? State the null and alternate hypotheses. H0: μ = 40; H1: μ ≠ 40 H0: p = 40; H1: p > 40 H0: μ = 40; H1: μ > 40 H0: μ = 40; H1: μ < 40 H0: p = 40; H1: p < 40 H0: p = 40; H1: p ≠ 40 (b) What sampling distribution will you use? What assumptions are you making? The standard normal, since we assume that x has a normal distribution with known σ. The Student's t, since we assume that x has a normal distribution with unknown σ. The Student's t, since we assume that x has a normal distribution with known σ. The standard normal, since we assume that x has a normal distribution with unknown σ. What is the value of the sample test statistic? (Round your answer to two decimal places.) (c) Find (or estimate) the P-value. P-value > 0.250 0.125 < P-value < 0.250 0.050 < P-value < 0.125 0.025 < P-value < 0.050 0.005 < P-value < 0.025 P-value < 0.005 Sketch the sampling distribution and show the area corresponding to the P-value. WebAssign Plot WebAssign Plot WebAssign Plot WebAssign Plot (d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis? Are the data statistically significant at level α? At the α = 0.01 level, we reject the null hypothesis and conclude the data are statistically significant. At the α = 0.01 level, we reject the null hypothesis and conclude the data are not statistically significant. At the α = 0.01 level, we fail to reject the null hypothesis and conclude the data are statistically significant. At the α = 0.01 level, we fail to reject the null hypothesis and conclude the data are not statistically significant. (e) Interpret your conclusion in the context of the application. There is sufficient evidence at the 0.01 level to conclude that the average number of matches per box is now greater than 40. There is insufficient evidence at the 0.01 level to conclude that the average number of matches per box is now greater than 40.
In: Statistics and Probability
The Nero Match Company sells matchboxes that are supposed to
have an average of 40 matches per box, with σ = 8. A
random sample of 98 matchboxes shows the average number of matches
per box to be 42.4. Using a 1% level of significance, can you say
that the average number of matches per box is more than 40?
What are we testing in this problem?
single mean single proportion
(a) What is the level of significance?
State the null and alternate hypotheses.
H0: p = 40; H1: p < 40H0: μ = 40; H1: μ ≠ 40 H0: p = 40; H1: p ≠ 40H0: μ = 40; H1: μ < 40H0: p = 40; H1: p > 40H0: μ = 40; H1: μ > 40
(b) What sampling distribution will you use? What assumptions are
you making?
The Student's t, since we assume that x has a normal distribution with known σ.The standard normal, since we assume that x has a normal distribution with unknown σ. The Student's t, since we assume that x has a normal distribution with unknown σ.The standard normal, since we assume that x has a normal distribution with known σ.
What is the value of the sample test statistic? (Round your answer
to two decimal places.)
(c) Find (or estimate) the P-value.
P-value > 0.2500.125 < P-value < 0.250 0.050 < P-value < 0.1250.025 < P-value < 0.0500.005 < P-value < 0.025P-value < 0.005
Sketch the sampling distribution and show the area corresponding to
the P-value.
(d) Based on your answers in parts (a) to (c), will you reject or
fail to reject the null hypothesis? Are the data statistically
significant at level α?
At the α = 0.01 level, we reject the null hypothesis and conclude the data are statistically significant.At the α = 0.01 level, we reject the null hypothesis and conclude the data are not statistically significant. At the α = 0.01 level, we fail to reject the null hypothesis and conclude the data are statistically significant.At the α = 0.01 level, we fail to reject the null hypothesis and conclude the data are not statistically significant.
(e) Interpret your conclusion in the context of the
application.
There is sufficient evidence at the 0.01 level to conclude that the average number of matches per box is now greater than 40.There is insufficient evidence at the 0.01 level to conclude that the average number of matches per box is now greater than 40.
In: Statistics and Probability
Based on what you observe here, would you recommend changing or keeping the warranty that goes along with this type of tire? If you change it, what would you recommend it to be? You must explain what are the possible pros and cons of your recommendations? While you don't need to do a financial analysis, present some of the costs that you considered in making your decisions. A tire company is testing their mileage warranty of 30,000 miles.
|
Tire # |
Mileage |
|
1 |
30564 |
|
2 |
28886 |
|
3 |
29822 |
|
4 |
29326 |
|
5 |
31535 |
|
6 |
31880 |
|
7 |
28376 |
|
8 |
29664 |
|
9 |
25001 |
|
10 |
31981 |
|
11 |
31048 |
|
12 |
29672 |
|
13 |
33267 |
|
14 |
31512 |
|
15 |
35228 |
|
16 |
27089 |
|
17 |
31372 |
|
18 |
32749 |
|
19 |
28670 |
|
20 |
31683 |
|
21 |
34551 |
|
22 |
31811 |
|
23 |
29535 |
|
24 |
25741 |
|
25 |
28874 |
|
26 |
28535 |
|
27 |
28444 |
|
28 |
27504 |
|
29 |
32179 |
|
30 |
31491 |
|
31 |
29531 |
|
32 |
29715 |
|
33 |
30103 |
|
34 |
33841 |
|
35 |
28462 |
|
36 |
29535 |
|
37 |
34548 |
|
38 |
28165 |
|
39 |
30817 |
|
40 |
34027 |
|
41 |
31802 |
|
42 |
31878 |
|
43 |
31972 |
|
44 |
30650 |
|
45 |
31514 |
|
46 |
28820 |
|
47 |
32669 |
|
48 |
33534 |
|
49 |
30598 |
|
50 |
29459 |
In: Statistics and Probability
A newly installed
automatic gate system was being tested to see if the number of
failures in 1,000 entry attempts was the same as the number of
failures in 1,000 exit attempts. A random sample of eight delivery
trucks was selected for data collection. Do these sample results
show that there is a significant difference between entry and exit
gate failures? Use α = 0.05.
| Truck 1 | Truck 2 | Truck 3 | Truck 4 | Truck 5 | Truck 6 | Truck 7 | Truck 8 | |
| Entry failures | 40 | 45 | 53 | 55 | 64 | 54 | 49 | 44 |
| Exit failures | 50 | 53 | 58 | 60 | 59 | 46 | 54 | 45 |
(a) Choose the appropriate hypotheses. Define the
difference as Entry − Exit.
H0: μd = 0 versus H1: μd ≠ 0
H0: μd ≥ 0 versus H1: μd < 0
H0: μd ≤ 0 versus H1: μd > 0
(b) Find the test statistic
tcalc. (Round your answer to 3 decimal
places. A negative value should be indicated by a minus
sign.)
tcalc ?
(c) Find the critical value
tcalc for α = 0.05. (Round
your answer to 3 decimal places. A negative value should be
indicated by a minus sign.)
tcalc?
(d) Find the p-value. (Round your
answer to 4 decimal places.)
What is the p-value?
In: Statistics and Probability
Television advertisers base their investment decisions regarding the promotion of their products and services on demographic information about television viewers. The age of the viewers is a key factor in their process. The following table shows the number of hours that a random sample of individuals watched television during the week. The individuals are grouped according to their ages. Minitab is required. You will need to enter the data into Minitab.
|
Age Group |
|||
|
18-24 |
25-34 |
35-49 |
50-64 |
|
49 |
41 |
44 |
39 |
|
33 |
40 |
19 |
14 |
|
33 |
33 |
27 |
15 |
|
39 |
35 |
36 |
17 |
|
71 |
21 |
49 |
20 |
a. At the 0.05 level of significance, determine if there is a difference in the mean number of hours of television watched by age group. State your hypotheses and show all 7 steps clearly. (14 points)
b. Give and interpret the p-value. (3 points)
c. Should Tukey pairwise comparisons be conducted? Why or why not? (3 points)
d. If appropriate, use Minitab to produce Tukey pairwise comparison. Write a few sentences with your conclusions from those comparisons. (4 points)
e. Use Levene’s test to determine if the assumption of homogeneity of variances is valid. Give the hypotheses, test statistic, p-value, decision, and conclusion. Use the 0.05 level of significance. (8 points)
f. Verify with Minitab by attaching or including relevant output. (6 points)
In: Math
Diamond and Turf Inc. is considering an investment in one of two machines. The sewing machine will increase productivity from sewing 180 baseballs per hour to sewing 324 per hour. The contribution margin per unit is $0.5 per baseball. Assume that any increased production of baseballs can be sold. The second machine is an automatic packing machine for the golf ball line. The packing machine will reduce packing labor cost. The labor cost saved is equivalent to $21 per hour. The sewing machine will cost $386,400, will have a seven-year life, and will operate for 1,400 hours per year. The packing machine will cost $94,300, will have a seven-year life, and will operate for 1,200 hours per year. Diamond and Turf seeks a minimum rate of return of 12% on its investments.
| Present Value of an Annuity of $1 at Compound Interest | |||||
| Year | 6% | 10% | 12% | 15% | 20% |
| 1 | 0.943 | 0.909 | 0.893 | 0.870 | 0.833 |
| 2 | 1.833 | 1.736 | 1.690 | 1.626 | 1.528 |
| 3 | 2.673 | 2.487 | 2.402 | 2.283 | 2.106 |
| 4 | 3.465 | 3.170 | 3.037 | 2.855 | 2.589 |
| 5 | 4.212 | 3.791 | 3.605 | 3.353 | 2.991 |
| 6 | 4.917 | 4.355 | 4.111 | 3.785 | 3.326 |
| 7 | 5.582 | 4.868 | 4.564 | 4.160 | 3.605 |
| 8 | 6.210 | 5.335 | 4.968 | 4.487 | 3.837 |
| 9 | 6.802 | 5.759 | 5.328 | 4.772 | 4.031 |
| 10 | 7.360 | 6.145 | 5.650 | 5.019 | 4.192 |
a. Determine the net present value for the two machines. Use the table of present values of an annuity of $1 above. Round to the nearest dollar.
| Sewing Machine | Packing Machine | |
| Present value of annual net cash flows | $ | $ |
| Amount to be invested | $ | $ |
| Net present value | $ | $ |
b. Determine the present value index for the two machines. If required, round your answers to two decimal places.
| Sewing Machine | Packing Machine | |
| Present value index |
2.
Bi-Coastal Railroad Inc. is considering acquiring equipment at a cost of $104,000. The equipment has an estimated life of 10 years and no residual value. It is expected to provide yearly net cash flows of $52,000. The company’s minimum desired rate of return for net present value analysis is 15%.
| Present Value of an Annuity of $1 at Compound Interest | |||||
| Year | 6% | 10% | 12% | 15% | 20% |
| 1 | 0.943 | 0.909 | 0.893 | 0.870 | 0.833 |
| 2 | 1.833 | 1.736 | 1.690 | 1.626 | 1.528 |
| 3 | 2.673 | 2.487 | 2.402 | 2.283 | 2.106 |
| 4 | 3.465 | 3.170 | 3.037 | 2.855 | 2.589 |
| 5 | 4.212 | 3.791 | 3.605 | 3.353 | 2.991 |
| 6 | 4.917 | 4.355 | 4.111 | 3.785 | 3.326 |
| 7 | 5.582 | 4.868 | 4.564 | 4.160 | 3.605 |
| 8 | 6.210 | 5.335 | 4.968 | 4.487 | 3.837 |
| 9 | 6.802 | 5.759 | 5.328 | 4.772 | 4.031 |
| 10 | 7.360 | 6.145 | 5.650 | 5.019 | 4.192 |
Compute the following:
a. The average rate of return, assuming the
annual earnings are equal to the net cash flows less the annual
depreciation expense on the equipment. If required, round your
answer to one decimal place.
_________%
b. The cash payback period.
_________years
c. The net present value. Use the above table of the present value of an annuity of $1. Round to the nearest dollar. If required, use a minus sign to indicate negative net present value" for current grading purpose.
| Present value of annual net cash flows | $____________ |
| Less amount to be invested | $____________ |
| Net present value | $____________ |
Thank you so much!!
In: Accounting
In a consumer study investigating preferences for a new packaging material for granola bars, with the objective of the test being the comparison between the new packaging material (Package A) and the current packaging material (B). A same-difference test was conducted using 180 randomly-selected consumers at a supermarket. Each consumer was presented with two samples, with the combinations being AA, AB, BA or BB. The following results were obtained. A worked example is in your text book (pg. 103) (7 pts)
|
Subject Received |
Subject Received |
|||
|
Matched Pair (AA, BB) |
Unmatched Pair (AB,BA) |
TOTAL |
||
|
Subject Said |
Same |
65 |
32 |
|
|
Different |
25 |
58 |
||
|
Total |
Questions:
Question 1: When panelists were presented with the same sample, how many panelists said the samples were the same? (ie. they were correct):
Question 2: When panelists were presented with the different sample, how many panelists said the samples were the same? (ie. they were incorrect):
Question 3: What is the expected total for the “different” condition?
Question 4: What is the calculated value for the same/match condition?
Question 5: What is the calculated value for the different/unmatched condition?
Question 6: What is TOTAL calculated chi-square value?
Question 7: For p=0.05, what is our critical value (taken from Table 19-5)? (0.5 pts)
Question 8: Are there significant differences between these two packaging materials? (0.5 pts)
In: Statistics and Probability