A company that sells baseball cards claims that 40% of the cards are rookies, 50% are veterans, and 10% are all stars. Suppose a random sample of 100 cards has 55 rookies, 40 veterans, and 5 all-stars.
|
Card Type |
Observed Frequency O |
Expected Frequency E |
O-E |
(O-E)2 |
(O-E)2 E |
|
Rookie |
|||||
|
Veteran |
|||||
|
All-Star |
Test the company's claim using a 0.01 level of significance. You may use the empty columns of the table above to assist you.
a) State the null and alternative hypothesis
b) Find the value of the test statistic and the critical value. You may use the empty columns of the table above to assist you.
c) State whether you should reject or fail to reject the null hypothesis. Justify your answer.
d) State you conclusion in non-technical terms
Please fill in the table with corresponding values and show all work.
In: Statistics and Probability
The current price of Gringotts Bank Corporation is $50. The price will increase by 40% or fall by 35% during each of the next two years. The company will pay a $9 dividend at the end of the first year if the stock price has risen, and will pay a $4 dividend if the price has fallen. It will not pay any dividends at the end of second year. The annualized, continuously compounded interest rate is 5%. What is the value of a 2-year European call option with a $45 strike price? What if it’s an American call option?
In: Finance
The following time series shows the sales of a particular product over the past 12 months.
| Month | Sales |
|---|---|
| 1 | 105 |
| 2 | 135 |
| 3 | 120 |
| 4 | 105 |
| 5 | 90 |
| 6 | 120 |
| 7 | 145 |
| 8 | 140 |
| 9 | 100 |
| 10 | 80 |
| 11 | 100 |
| 12 | 110 |
(b)
Use α = 0.4 to compute the exponential smoothing forecasts for the time series. (Round your answers to two decimal places.)
| Month t | Time Series Value
Yt |
Forecast
Ft |
|---|---|---|
| 1 | 105 | |
| 2 | 135 | |
| 3 | 120 | |
| 4 | 105 | |
| 5 | 90 | |
| 6 | 120 | |
| 7 | 145 | |
| 8 | 140 | |
| 9 | 100 | |
| 10 | 80 | |
| 11 | 100 | |
| 12 | 110 |
(c)
Use a smoothing constant of α = 0.6 to compute the exponential smoothing forecasts. (Round your answers to two decimal places.)
| Month t | Time Series Value
Yt |
Forecast
Ft |
|---|---|---|
| 1 | 105 | |
| 2 | 135 | |
| 3 | 120 | |
| 4 | 105 | |
| 5 | 90 | |
| 6 | 120 | |
| 7 | 145 | |
| 8 | 140 | |
| 9 | 100 | |
| 10 | 80 | |
| 11 | 100 | |
| 12 | 110 |
In: Statistics and Probability
write C# console app No arrays
Credit card validation Problem. A credit card number must be between 13 and 16 digits. It must start with: –4 for visa cards –5 for mater cards –37 for American express cards –6 for discover cards •Consider a credit card number 4380556218442727 •Step1: double every second digit from right to left. If doubling of a digit results in a two-digit number, add up the two digits to get a single-digit number. •Step2: now add all the single digit numbers from step1 4 + 4 + 8 + 2 + 3 + 1 + 7 + 8 •Step3: add all digits in the odd places from right to left in the card number 7+ 7 + 4 + 8 + 2 + 5 + 0 + 3 = 36 •Step4: sum the results from step 2 and step 3 37 + 36 = 73 •Step 5: if the results from step 4 is divisible by 10, the card number is valid; otherwise, it is invalid. In this case, the card number is not valid – because 75 not divisible by 10. use modulus operator
In: Computer Science
Question 3
A brain specialist performed an experiment to see if sensory deprivation over an extended period of time reduces the alpha-wave frequencies produced by the brain. To determine this, 10 subjects, inmates of a Correctional Centre, were randomly selected.
First, the members of the group were allowed to remain in their cells. Seven days later, alpha-wave frequencies were measured in Hertz for all subjects (nonconfined).
After that, the same group of subjects were placed in solitary confinement for seven days and their alpha-wave frequencies were measured again in Hertz.
The data from both confined and nonconfined groups of the experiment are recorded in the following table.
|
Subject # |
Confined |
Nonconfined |
|
1 |
7 |
8 |
|
2 |
10 |
9 |
|
3 |
9 |
10 |
|
4 |
9 |
10 |
|
5 |
10 |
11 |
|
6 |
9 |
11 |
|
7 |
11 |
11 |
|
8 |
12 |
11 |
|
9 |
11 |
12 |
|
10 |
12 |
14 |
Use a chi-square test to investigate if the solitary confinement reduces the alpha-wave frequencies of the brain. In performing this test, state appropriate hypothesis (define any symbols used). Calculate the value of the test statistic, P-value of this test and write a meaningful conclusion.
(While a paired t-test is what would usually be used for this question, we are specifically asking you to perform a chi-square test).
In: Statistics and Probability
Hassad owns a rental house on Lake Tahoe. He uses a real estate firm to screen prospective renters, but he makes the final decision on all rentals. He also is responsi- ble for setting the weekly rental price of the house. During the current year, the house rents for $1,500 per week. Hassad pays a commission of $150 and a cleaning fee of
$75 for each week the property is rented. During the current year, he incurs the follow- ing additional expenses related to the property:
|
Mortgage interest |
$12,000 |
|
Property taxes |
2,700 |
|
Utilities |
1,400 |
|
Landscaping fees |
900 |
|
Repairs |
450 |
|
Depreciation (unallocated) |
7,500 |
a. What is the proper tax treatment if Hassad rents the house for only 1 week (7 days) and uses it 50 days for personal purposes?
b. What is the proper tax treatment if Hassad rents the house for 8 weeks (56 days) and uses it 44 days for personal purposes?
c. What is the proper tax treatment if Hassad rents the house for 25 weeks (175 days) and uses it 15 days for personal purposes?
In: Accounting
1. The effectiveness of a blood-pressure drug is being
investigated. An experimenter finds that, on average, the reduction
in systolic blood pressure is 44 for a sample of size 22 and
standard deviation 6. Estimate how much the drug will lower a
typical patient's systolic blood pressure (using a 95% confidence
level).
Give your answers to one decimal place and provide the point
estimate with its margin of error. __________________ ±
________________________
2. In a survey, 31 people were asked how much they spent on
their child's last birthday gift. The results were roughly
bell-shaped with a mean of $42 and standard deviation of $10.
Estimate how much a typical parent would spend on their child's
birthday gift (use a 95% confidence level).
Give your answers to one decimal place. Provide the point estimate
and margin or error. _______________ ± _________________
3. You must estimate the mean temperature (in degrees Fahrenheit) with the following sample temperatures:
| 13.8 |
| 65.2 |
| 51.2 |
| 22.5 |
| 41.7 |
| 13.8 |
| 58.4 |
| 39.4 |
| 32.2 |
| 31.1 |
Find the 80% confidence interval. Enter your answer as an
open-interval (i.e., parentheses)
accurate to two decimal places (because the sample data are
reported accurate to one decimal place). 80% C.I. =
___________________
Answer should be obtained without any preliminary rounding.
4. The effectiveness of a blood-pressure drug is being
investigated. An experimenter finds that, on average, the reduction
in systolic blood pressure is 38.9 for a sample of size 339 and
standard deviation 13.4. Estimate how much the drug will lower a
typical patient's systolic blood pressure (using a 80% confidence
level).
Enter your answer as a tri-linear inequality accurate to one
decimal place (because the sample statistics are reported accurate
to one decimal place).
_________< μ < __________
5. Assume that a sample is used to estimate a population mean
μμ. Find the margin of error M.E. that corresponds to a
sample of size 7 with a mean of 52.6 and a standard deviation of
14.3 at a confidence level of 99.9%. Report ME accurate to one
decimal place because the sample statistics are presented with this
accuracy.
M.E. = ________________ Answer should be
obtained without any preliminary rounding. However, the critical
value may be rounded to 3 decimal places.
6. Express the confidence interval (337.6,540.6)(337.6,540.6) in
the form of ¯x ± ME.
¯x ± ME= ___________ + ______________
7. Assume that a sample is used to estimate a population mean
μμ. Find the 95% confidence interval for a sample of size 66 with a
mean of 27.9 and a standard deviation of 20.8. Enter your answer as
an open-interval (i.e., parentheses)
accurate to one decimal place (because the sample statistics are
reported accurate to one decimal place).
95% C.I. = _____________
Answer should be obtained without any preliminary rounding.
However, the critical value may be rounded to 3 decimal places.
8. Assume that a sample is used to estimate a population
proportion μμ. Find the margin of error M.E. that
corresponds to a sample of size 411 with a mean of 20.9 and a
standard deviation of 20.6 at a confidence level of 90%.
Report ME accurate to one decimal place because the sample
statistics are presented with this accuracy.
M.E. =__________________
Answer should be obtained without any preliminary rounding.
However, the critical value may be rounded to 3 decimal places.
9. In a survey, 16 people were asked how much they spent on
their child's last birthday gift. The results were roughly
bell-shaped with a mean of $38 and standard deviation of $7.
Estimate how much a typical parent would spend on their child's
birthday gift (use a 95% confidence level).
Give your answers to one decimal place. Provide the point estimate
and margin or error. _____________ ± _______________
10. Assume that a sample is used to estimate a population mean
μμ. Find the 90% confidence interval for a sample of size 56 with a
mean of 59.2 and a standard deviation of 13.2. Enter your answer as
an open-interval (i.e., parentheses)
accurate to one decimal place (because the sample statistics are
reported accurate to one decimal place).
90% C.I. = _________________
Answer should be obtained without any preliminary rounding.
However, the critical value may be rounded to 3 decimal places.
11. Assume that a sample is used to estimate a population mean
μμ. Find the margin of error M.E. that corresponds to a
sample of size 21 with a mean of 58.7 and a standard deviation of
17.9 at a confidence level of 80%.
Report ME accurate to one decimal place because the sample
statistics are presented with this accuracy.
M.E. = _______________
Answer should be obtained without any preliminary rounding.
However, the critical value may be rounded to 3 decimal places.
12. Express the confidence interval (174.8,273.6) in the form of ¯x ± ME.
¯x ± ME=_________________
13. Assume that a sample is used to estimate a population
proportion μμ. Find the margin of error M.E. that
corresponds to a sample of size 45 with a mean of 15.3 and a
standard deviation of 6.6 at a confidence level of 99%.
Report ME accurate to one decimal place because the sample
statistics are presented with this accuracy.
M.E. = ________________
Answer should be obtained without any preliminary rounding.
However, the critical value may be rounded to 3 decimal places.
14. You must estimate the mean temperature (in degrees Fahrenheit) with the following sample temperatures:
| 32.1 |
| 30.7 |
| 9 |
| 22.2 |
| 42.7 |
| -5.8 |
| 20.5 |
| 20.8 |
| 52.4 |
| 28.6 |
| 9.2 |
Find the 90% confidence interval. Enter your answer as an
open-interval (i.e., parentheses)
accurate to two decimal places (because the sample data are
reported accurate to one decimal place).
90% C.I. = _________________
15. The effectiveness of a blood-pressure drug is being
investigated. An experimenter finds that, on average, the reduction
in systolic blood pressure is 30.9 for a sample of size 889 and
standard deviation 11.3. Estimate how much the drug will lower a
typical patient's systolic blood pressure (using a 95% confidence
level).
Enter your answer as a tri-linear inequality accurate to one
decimal place (because the sample statistics are reported accurate
to one decimal place).
___________< μ < ____________________
16. You must estimate the mean temperature (in degrees Fahrenheit) with the following sample temperatures:
| 74.2 |
| 89.8 |
| 70.1 |
| 81.7 |
| 63.6 |
| 94.8 |
| 75.3 |
| 73.3 |
| 76 |
| 66.3 |
| 63 |
Find the 80% confidence interval. Enter your answer as an open-interval (i.e., parentheses) accurate to two decimal places (because the sample data are reported accurate to one decimal place). 80% C.I. = _____________________
17. The effectiveness of a blood-pressure drug is being
investigated. An experimenter finds that, on average, the reduction
in systolic blood pressure is 50 for a sample of size 19 and
standard deviation 8. Estimate how much the drug will lower a
typical patient's systolic blood pressure (using a 90% confidence
level).
Give your answers to one decimal place and provide the point
estimate with its margin of error.
______________ ± _______________
In: Math
A set of solar batteries is used in a research satellite. The
satellite can run on only one battery, but it runs best if more
than one battery is used. The variance σ2 of lifetimes of these
batteries affects the useful lifetime of the satellite before it
goes dead. If the variance is too small, all the batteries will
tend to die at once. Why? If the variance is too large, the
batteries are simply not dependable. Why? Engineers have determined
that a variance of σ2 = 23 months (squared) is most desirable for
these batteries. A random sample of 30 batteries gave a sample
variance of 14.6 months (squared). Using a 0.05 level of
significance, test the claim that σ2 = 23 against the claim that σ2
is different from 23.
(a) What is the level of significance?
State the null and alternate hypotheses.
Ho: σ2 = 23; H1: σ2 > 23
Ho: σ2 = 23; H1: σ2 ≠ 23
Ho: σ2 > 23; H1: σ2 = 23
Ho: σ2 = 23; H1: σ2 < 23
(b) Find the value of the chi-square statistic for the sample. (Round your answer to two decimal places.)
What are the degrees of freedom?
What assumptions are you making about the original
distribution?
We assume a uniform population distribution.
We assume a exponential population distribution.
We assume a normal population distribution.
We assume a binomial population distribution.
(c) Find or estimate the P-value of the sample test
statistic.
P-value > 0.100
0.050 < P-value < 0.100
0.025 < P-value < 0.050
0.010 < P-value < 0.025
0.005 < P-value < 0.010
P-value < 0.005
(d) Based on your answers in parts (a) to (c), will you reject
or fail to reject the null hypothesis?
Since the P-value > α, we fail to reject the null
hypothesis.
Since the P-value > α, we reject the null hypothesis.
Since the P-value ≤ α, we reject the null hypothesis.
Since the P-value ≤ α, we fail to reject the null hypothesis.
(e) Interpret your conclusion in the context of the
application.
At the 5% level of significance, there is insufficient evidence to
conclude that the variance of battery life is different from
23.
At the 5% level of significance, there is sufficient evidence to
conclude that the variance of battery life is different from
23.
(f) Find a 90% confidence interval for the population variance.
(Round your answers to two decimal places.)
lower limit
upper limit
(g) Find a 90% confidence interval for the population standard
deviation. (Round your answers to two decimal places.)
lower limit
months
upper limit
months
In: Statistics and Probability
A set of solar batteries is used in a research satellite. The satellite can run on only one battery, but it runs best if more than one battery is used. The variance σ2 of lifetimes of these batteries affects the useful lifetime of the satellite before it goes dead. If the variance is too small, all the batteries will tend to die at once. Why? If the variance is too large, the batteries are simply not dependable. Why? Engineers have determined that a variance of σ2 = 23 months (squared) is most desirable for these batteries. A random sample of 26 batteries gave a sample variance of 15.2 months (squared). Using a 0.05 level of significance, test the claim that σ2 = 23 against the claim that σ2 is different from 23.
(a) What is the level of significance?
State the null and alternate hypotheses.
Ho: σ2 = 23; H1: σ2 > 23Ho: σ2 = 23; H1: σ2 ≠ 23 Ho: σ2 > 23; H1: σ2 = 23Ho: σ2 = 23; H1: σ2 < 23
(b) Find the value of the chi-square statistic for the sample.
(Round your answer to two decimal places.)
What are the degrees of freedom?
What assumptions are you making about the original
distribution?
We assume a exponential population distribution.We assume a binomial population distribution. We assume a normal population distribution.We assume a uniform population distribution.
(c) Find or estimate the P-value of the sample test
statistic.
P-value > 0.1000.050 < P-value < 0.100 0.025 < P-value < 0.0500.010 < P-value < 0.0250.005 < P-value < 0.010P-value < 0.005
(d) Based on your answers in parts (a) to (c), will you reject or
fail to reject the null hypothesis?
Since the P-value > α, we fail to reject the null hypothesis.Since the P-value > α, we reject the null hypothesis. Since the P-value ≤ α, we reject the null hypothesis.Since the P-value ≤ α, we fail to reject the null hypothesis.
(e) Interpret your conclusion in the context of the
application.
At the 5% level of significance, there is insufficient evidence to conclude that the variance of battery life is different from 23.At the 5% level of significance, there is sufficient evidence to conclude that the variance of battery life is different from 23.
(f) Find a 90% confidence interval for the population variance.
(Round your answers to two decimal places.)
| lower limit | |
| upper limit |
(g) Find a 90% confidence interval for the population standard
deviation. (Round your answers to two decimal places.)
| lower limit | months |
| upper limit | months |
In: Statistics and Probability
A set of solar batteries is used in a research satellite. The satellite can run on only one battery, but it runs best if more than one battery is used. The variance σ2 of lifetimes of these batteries affects the useful lifetime of the satellite before it goes dead. If the variance is too small, all the batteries will tend to die at once. Why? If the variance is too large, the batteries are simply not dependable. Why? Engineers have determined that a variance of σ2 = 23 months (squared) is most desirable for these batteries. A random sample of 30 batteries gave a sample variance of 14.4 months (squared). Using a 0.05 level of significance, test the claim that σ2 = 23 against the claim that σ2 is different from 23.
(a) What is the level of significance?
State the null and alternate hypotheses.
Ho: σ2 = 23; H1: σ2 > 23Ho: σ2 = 23; H1: σ2 < 23 Ho: σ2 > 23; H1: σ2 = 23Ho: σ2 = 23; H1: σ2 ≠ 23
(b) Find the value of the chi-square statistic for the sample.
(Round your answer to two decimal places.)
What are the degrees of freedom?
What assumptions are you making about the original
distribution?
We assume a normal population distribution.We assume a uniform population distribution. We assume a exponential population distribution.We assume a binomial population distribution.
(c) Find or estimate the P-value of the sample test
statistic.
P-value > 0.1000.050 < P-value < 0.100 0.025 < P-value < 0.0500.010 < P-value < 0.0250.005 < P-value < 0.010P-value < 0.005
(d) Based on your answers in parts (a) to (c), will you reject or
fail to reject the null hypothesis?
Since the P-value > α, we fail to reject the null hypothesis.Since the P-value > α, we reject the null hypothesis. Since the P-value ≤ α, we reject the null hypothesis.Since the P-value ≤ α, we fail to reject the null hypothesis.
(e) Interpret your conclusion in the context of the
application.
At the 5% level of significance, there is insufficient evidence to conclude that the variance of battery life is different from 23.At the 5% level of significance, there is sufficient evidence to conclude that the variance of battery life is different from 23.
(f) Find a 90% confidence interval for the population variance.
(Round your answers to two decimal places.)
| lower limit | |
| upper limit |
(g) Find a 90% confidence interval for the population standard
deviation. (Round your answers to two decimal places.)
| lower limit | months |
| upper limit | months |
In: Math