3. An engineer wanted to determine how the weight of a car (kilograms), ?, affects fuel consumption (kilometers per litre), ?. The weights and fuel consumption for a random sample of 10 cars were recorded. Refer to the regression results from Excel.
Regression Statistics
Multiple R |
0.84157604 |
R Square |
0.708250231 |
Adjusted R Square |
0.671781509 |
Standard Error |
0.712010044 |
Observations |
10 |
Coefficients |
Standard Error |
t-Stat |
P-Value |
Lower 95% |
Upper 95% |
|
Intercept |
15.883033 |
1.680071 |
9.453785 |
0.000013 |
12.008782 |
19.757285 |
X-Variable 1 |
-0.004384 |
0.000995 |
(?) |
0.002266 |
-0.006679 |
-0.002090 |
a) Write the equation of the estimated least-squares regression line.
b) Test whether a linear relationship exists between
weight of a car and fuel consumption at the
? = 0.01 level of significance. Use the critical value
method.
c) Use the Excel output to write a 95% confidence interval for the slope of the true least-squares regression line. Interpret the result. Write a descriptive statement that could be easily understood by someone who is not in this statistics class.
d) A 95% prediction interval for ? = 1900 is 5.83 to
9.28. Interpret this result. Write a descriptive
statement that could be easily understood by someone who is not in
this statistics class.
In: Statistics and Probability
Cynthia is a rideshare driver, and she often starts and stops her car. Each time Cynthia starts her car it either runs or stalls. If her car stalls when she starts it, then it always runs the next time she starts it. If her car runs when she starts it, then next time it is ten times as likely to stall as it is to run.
How often in the long run does Cynthia's car run when she starts it? Answer with a fraction in lowest terms.
In: Statistics and Probability
Calculate each binomial probability:
(a) Fewer than 5 successes in 11 trials with a 10
percent chance of success. (Round your answer to 4 decimal
places.)
Probability =
(b) At least 2 successes in 8 trials with a 30
percent chance of success. (Round your answer to 4 decimal
places.)
Probability =
(c) At most 10 successes in 18 trials with a 70
percent chance of success. (Round your answer to 4 decimal
places.)
Probability =
In: Statistics and Probability
A person has a net asset of $1 million, including a $300,000 net equity of a house (market value of the house – mortgage). Specifically, the house has a market value of $500,000 including $300,000 for the structure and $200,000 for the land, and a mortgage of $200,000. The person plans to buy $300,000 fire insurance for full coverage of the house. For simplicity, assume that each year the house has a 1% probability of being totally destroyed by fire and a 99% probability of no damage occurring to the house. The person’s utility for money is approximately proportional to the quartic root of money with U($100,000,000)=100 and U($0)=0.
7a.(5 points) Draw the decision tree for the person’s decision of buying or not buying the insurance.
7b.(10 points) Determine the maximum insurance premium IP the person would be willing to pay.
7c. (5 points) What is the risk premium at the maximum IP?
Determine the maximum insurance premium the person would be willing to pay for a $200,000 insurance just to cover the mortgage.
(Hint: in this case, the house is under-insured. In other words, with the $200,000 insurance, if the house is totally destroyed by fire, the person will suffer a loss in the net asset because the insurance covers only the mortgage not the full net equity of the house, and the maximum insurance premium the person would be willing to pay will need to be determined through numerical iterations).
In: Statistics and Probability
Sheets of aluminum from a supplier have a thickness that is normally distributed with a mean of 50 mm and a standard deviation of 4 mm (call this random variable X). Your company compresses the aluminum with a tool that is normally distributed with a mean of 20 mm and a standard deviation of 3 mm (call this random variable Y). You are interested in the random variable V = X – Y, the random variable V is the final aluminum thickness.
1. What is the probability that the outputted aluminum (that is, V), will be between 25 mm and 32 mm?
2. What is the probability that the outputted aluminum (that is, V), will be between 26.5 mm and 33.5 mm?
3. If the company had the choice between compressing aluminum to between 25-32 mm or 26.5-33.5 mm, then which is preferred (or neither)?
4. In one or two sentences, why is one preferred (if either) over the other [continuation of 3.3]; if neither are preferred, then why?
In: Statistics and Probability
Given the transition matrix P for a Markov chain, find the stable vector W. Write entries as fractions in lowest terms.
P= 0.5 0 0.5
0.2 0.2 0.6
0
1 0
In: Statistics and Probability
An orange juice producer buys oranges from a large orange grove that has one variety of orange. The amount of juice squeezed from these oranges is approximately normally distributed, with a mean of 4.40 ounces and a standard deviation of 0.32 ounce. Suppose that you select a sample of 16 oranges.
a. What is the probability that the sample mean amount of juice will be at least 4.27 ounces?
b. The probability is 72% that the sample mean amount of juice will be contained between what two values symmetrically distributed around the population mean?
c. The probability is 74% that the sample mean amount of juice will be greater than what value?
In: Statistics and Probability
What is a sampling distribution of means? How is it created and what is it used for?
In: Statistics and Probability
A recent study of 3100 children randomly selected found 20% of them deficient in vitamin D.
a) Construct the 98% confidence interval for the true proportion of children who are deficient in vitamin D.
left parenthesis nothing comma nothing right parenthesis,
(Round to three decimal places as needed.)
In: Statistics and Probability
A hospital conducted a study of the waiting time in its
emergency
room. The hospital has a main campus and three
satellite
locations. Management had a business objective of reducing
waiting time for emergency room cases that did not require
immediate attention. To study this, a random sample of 15
emergency
room cases that did not require immediate attention at each
location were selected on a particular day, and the waiting
times
(measured from check-in to when the patient was called into
the
clinic area) were collected and stored in ERWaiting .
a. At the 0.05 level of significance, is there evidence of a
difference
in the mean waiting times in the four locations?
Data Is Below:
Main | Satellite 1 | Satellite 2 | Satellite 3 |
120.08 | 30.75 | 75.86 | 54.05 |
81.90 | 61.83 | 37.88 | 38.82 |
78.79 | 26.40 | 68.73 | 36.85 |
63.83 | 53.84 | 51.08 | 32.83 |
79.77 | 72.30 | 50.21 | 52.94 |
47.94 | 53.09 | 58.47 | 34.13 |
79.88 | 27.67 | 86.29 | 69.37 |
48.63 | 52.46 | 62.90 | 78.52 |
55.43 | 10.64 | 44.84 | 55.95 |
64.06 | 53.50 | 64.17 | 49.61 |
64.99 | 37.28 | 50.68 | 66.40 |
53.82 | 34.31 | 47.97 | 76.06 |
62.43 | 66.00 | 60.57 | 11.37 |
65.07 | 8.99 | 58.37 | 83.51 |
81.02 | 29.75 | 30.40 | 39.17 |
In: Statistics and Probability
Measuring the height of a California redwood tree is very
difficult because these trees grow to heights of over 300 feet.
People
familiar with these trees understand that the height of a
California
redwood tree is related to other characteristics of the tree,
including the diameter of the tree at the breast height of a
person.
The data in Redwood represent the height (in feet) and
diameter
(in inches) at the breast height of a person for a sample of 21
California
redwood trees.
a. Assuming a linear relationship, use the least-squares
method
to compute the regression coefficients b0 and b1. State the
regression
equation that predicts the height of a tree based on the
tree’s diameter at breast height of a person.
b. Interpret the meaning of the slope in this equation.
c. Predict the mean height for a tree that has a breast height
diameter
of 25 inches.
d. Interpret the meaning of the coefficient of determination in
this
problem.
e. Perform a residual analysis on the results and determine the
adequacy
of the model.
Data Is Below:
Height | Diameter at breast height | Bark thickness |
122.0 | 20 | 1.1 |
193.5 | 36 | 2.8 |
166.5 | 18 | 2.0 |
82.0 | 10 | 1.2 |
133.5 | 21 | 2.0 |
156.0 | 29 | 1.4 |
172.5 | 51 | 1.8 |
81.0 | 11 | 1.1 |
148.0 | 26 | 2.5 |
113.0 | 12 | 1.5 |
84.0 | 13 | 1.4 |
164.0 | 40 | 2.3 |
203.3 | 52 | 2.0 |
174.0 | 30 | 2.5 |
159.0 | 22 | 3.0 |
205.0 | 42 | 2.6 |
223.5 | 45 | 4.3 |
195.0 | 54 | 4.0 |
232.5 | 39 | 2.2 |
190.5 | 36 | 3.5 |
100.0 | 8 | 1.4 |
In: Statistics and Probability
In: Statistics and Probability
In: Statistics and Probability
To properly treat patients, drugs prescribed by physicians must have a potency that is accurately defined. Consequently, not only must the distribution of potency values for shipments of a drug have a mean value as specified on the drug's container, but also the variation in potency must be small. Otherwise, pharmacists would be distributing drug prescriptions that could be harmfully potent or have a low potency and be ineffective. A drug manufacturer claims that its drug is marketed with a potency of 5 ± 0.1 milligram per cubic centimetre (mg/cc). A random sample of four containers gave potency readings equal to 4.93, 5.08, 5.03, and 4.89 mg/cc.
(a) Do the data present sufficient evidence to indicate that the
mean potency differs from 5 mg/cc? (Use α = 0.05. Round
your answers to three decimal places.)
1-2. Null and alternative hypotheses:
H0: μ = 5 versus Ha: μ < 5H0: μ ≠ 5 versus Ha: μ = 5 H0: μ = 5 versus Ha: μ > 5H0: μ < 5 versus Ha: μ > 5H0: μ = 5 versus Ha: μ ≠ 5
3. Test statistic: t
=
4. Rejection region: If the test is one-tailed, enter NONE for the
unused region.
t | > | |
t | < |
5. Conclusion:
H0 is not rejected. There is insufficient evidence to indicate that the mean potency differs from 5 mg/cc.H0 is not rejected. There is sufficient evidence to indicate that the mean potency differs from 5 mg/cc. H0 is rejected. There is sufficient evidence to indicate that the mean potency differs from 5 mg/cc.H0 is rejected. There is insufficient evidence to indicate that the mean potency differs from 5 mg/cc.
(b) Do the data present sufficient evidence to indicate that the
variation in potency differs from the error limits specified by the
manufacturer? (HINT: It is sometimes difficult to determine exactly
what is meant by limits on potency as specified by a manufacturer.
Since it implies that the potency values will fall into the
interval 5.0 ± 0.1 mg/cc with very high probability—the implication
is always—let us assume that the range 0.2; or (4.9 to
5.1), represents 6σ, as suggested by the Empirical Rule.
Note that letting the range equal 6σ rather than
4σ places a stringent interpretation on the manufacturer's
claim. We want the potency to fall into the interval
5.0 ± 0.1
with very high probability.) (Use α = 0.05. Round your
answers to three decimal places.)
1-2. Null and alternative hypotheses:
H0: σ2 = 0.0011 versus Ha: σ2 < 0.0011H0: σ2 > 0.0011 versus Ha: σ2 < 0.0011 H0: σ2 = 0.2 versus Ha: σ2 ≠ 0.2H0: σ2 = 0.0011 versus Ha: σ2 > 0.0011H0: σ2 = 0.2 versus Ha: σ2 > 0.2
3. Test statistic: χ2
=
4. Rejection region: If the test is one-tailed, enter NONE for the
unused region.
χ2 > |
χ2 < |
5. Conclusion:
H0 is rejected. There is insufficient evidence to indicate that the variation in potency differs from the specified error limits.H0 is rejected. There is sufficient evidence to indicate that the variation in potency differs from the specified error limits. H0 is not rejected. There is insufficient evidence to indicate that the variation in potency differs from the specified error limits.H0 is not rejected. There is sufficient evidence to indicate that the variation in potency differs from the specified error limits.
In: Statistics and Probability
You are a public health nurse that is interested in comparing the total blood cholesterol levels in clients from different census tracts in the city that are enrolled in a community-based lifestyle program offer by the Public Health Department. Previous research has shown that cholesterol levels are generally normally distributed, and you have no reason to believe that the variances in cholesterol levels would differ across census tracts. The table below provides the data that you collected on cholesterol levels in clients enrolled in your program from four census tracts in the city. Assuming a 0.05 level of significance, can you conclude that cholesterol levels differ across the census tracts? If there is a difference, which census tracts differ from the others?
Area A |
Area B |
Area C |
Area D |
96 |
120 |
138 |
101 |
126 |
122 |
139 |
104 |
166 |
128 |
140 |
104 |
168 |
131 |
140 |
107 |
173 |
132 |
141 |
108 |
178 |
147 |
159 |
122 |
190 |
148 |
166 |
125 |
194 |
149 |
172 |
136 |
195 |
151 |
176 |
139 |
198 |
160 |
180 |
149 |
212 |
160 |
182 |
160 |
212 |
162 |
185 |
162 |
213 |
164 |
188 |
162 |
215 |
165 |
188 |
168 |
216 |
174 |
192 |
171 |
227 |
184 |
194 |
173 |
185 |
194 |
174 |
|
188 |
200 |
175 |
|
192 |
202 |
180 |
|
194 |
203 |
182 |
|
197 |
206 |
182 |
|
198 |
210 |
184 |
|
201 |
215 |
184 |
|
221 |
220 |
190 |
|
224 |
222 |
191 |
|
230 |
224 |
192 |
|
242 |
230 |
193 |
|
250 |
231 |
194 |
|
251 |
232 |
195 |
|
233 |
200 |
||
237 |
206 |
||
243 |
207 |
||
245 |
207 |
||
263 |
209 |
||
274 |
210 |
||
294 |
212 |
||
307 |
215 |
||
327 |
221 |
||
224 |
|||
227 |
|||
229 |
|||
231 |
|||
239 |
|||
241 |
|||
248 |
|||
262 |
|||
317 |
In: Statistics and Probability