In: Math
For this problem, carry at least four digits after the decimal
in your calculations. Answers may vary slightly due to
rounding.
In a random sample of 63 professional actors, it was found that 39
were extroverts.
(a) Let p represent the proportion of all actors who
are extroverts. Find a point estimate for p. (Round your
answer to four decimal places.)
(b) Find a 95% confidence interval for p. (Round your
answers to two decimal places.)
lower limit | |
upper limit |
Give a brief interpretation of the meaning of the confidence
interval you have found.
We are 5% confident that the true proportion of actors who are extroverts falls within this interval.We are 95% confident that the true proportion of actors who are extroverts falls outside this interval. We are 5% confident that the true proportion of actors who are extroverts falls above this interval.We are 95% confident that the true proportion of actors who are extroverts falls within this interval.
(c) Do you think the conditions np > 5 and nq
> 5 are satisfied in this problem? Explain why this would be an
important consideration.
Yes, the conditions are satisfied. This is important because it allows us to say that p̂ is approximately binomial.No, the conditions are not satisfied. This is important because it allows us to say that p̂ is approximately normal. Yes, the conditions are satisfied. This is important because it allows us to say that p̂ is approximately normal.No, the conditions are not satisfied. This is important because it allows us to say that p̂ is approximately binomial.
In: Math
A study examined whether or not noses continue to grow throughout a person’s lifetime. The study included many measurements (including size of the nose as measured by total volume) and included multiple tests. For each of the tests described below:
a) State the null and alternative hypotheses.
b) Give a formal decision using a 5% significance level, and interpret the conclusion in context.
In: Math
The failure rate in a statistics class is 20%. In a class of 30 students, find the probability that exactly five students will fail. Use the normal distribution to approximate the binomial distribution.
In: Math
The mean number of pets per household is 2.96 with standard deviation 1.4. A sample of 52 households is drawn. Find the probability that the sample mean is less than 3.11.
a. |
0.2245 |
|
b. |
0.5676 |
|
c. 0.3254 |
||
d. |
0.7726 |
In: Math
The pH of 20 randomly selected lakes is measured. Their average pH is 5.7.
Part A. Historically the standard deviation in the pH values is
0.9. Use this standard deviation for the following questions.
Part Ai. Build a 95% confidence interval for the population mean
lake pH.
Part Aii. Build a 90% confidence interval for the population mean
lake pH.
Part Aiii. Build an 80% confidence interval for the population mean
lake pH.
Part Aiv. Compare the intervals you created in Ai, Aii and Aiii.
What effect does changing the level of confidence have on the
interval?
Please solve only part B
Part B. For the 20 measured lakes the standard deviation in the
pH values is 0.9. Use this standard deviation for the following
questions.
Part Bi. Build a 95% confidence interval for the population mean
lake pH.
Part Bii. Compare the confidence intervals in Bi and Ai. What
effect does not knowing the value of ? have on the interval?
Part Biii. Test whether the population mean pH differs from 6.
In: Math
For 300 trading days, the daily closing price of a stock (in $) is well modeled by a Normal model with a mean of
$197.12197.12 and a standard deviation of
$7.187.18. According to this model, what is the probability that on a randomly selected day in this period, the stock price closed as follows.
a) above $204.30204.30?
b) below $211.48211.48?
c) between $182.76182.76 and $211.48211.48?
In: Math
Hale's TV Productions is considering producing a pilot for a comedy series in the hope of selling it to a major television network. The network may decide to reject the series, but it may also decide to purchase the rights to the series for either one or two years. At this point in time, Hale may either produce the pilot and wait for the network's decision or transfer the rights for the pilot and series to a competitor for $250,000. Hale's decision alternatives and profits (in thousands of dollars) are as follows:
State of Nature | |||
Decision Alternative | Reject, S1 | 1 Year, S2 | 2 Years, S3 |
Produce pilot, d1 | -150 | 50 | 450 |
Sell to competitor, d2 | 250 | 250 | 250 |
The probabilities for the states of nature are P(S1) = 0.20, P(S2) = 0.30, and P(S3) = 0.50. For a consulting fee of $45,000, an agency will review the plans for the comedy series and indicate the overall chances of a favorable network reaction to the series. Assume that the agency review will result in a favorable (F) or an unfavorable (U) review and that the following probabilities are relevant:
P(F) = 0.63 | P(S1|F) = 0.07 | P(S1|U) = 0.41 |
P(U) = 0.37 | P(S2|F) = 0.27 | P(S2|U) = 0.38 |
P(S3|F) = 0.66 | P(S3|U) = 0.21 |
a. What is the expected value?
b. What is the expected value of perfect information?
c. What is the expected value of the agency's information? Round your answer to two decimal places.
d. What is the maximum that Hale should be willing to pay for the information? Round your answer to two decimal places.
In: Math
In each of parts (a)-(c), we have given a likely range for the observed value of a sample proportion p. Based on the given range, identify the educated guess that should be used for the observed value of p to calculate the required sample size for a prescribed confidence level and margin of error.
a. 0.2 to 0.3
b. 0.1 or less
c. 0.3 or greater
In: Math
Physical activity of obese young adults. In a study on the physical activity of young adults, pediatric researchers measured overall physical activity as the total number of registered movements (counts) over a period of time and then computed the number of counts per minute (cpm) for each subject (International Journal of Obesity, Jan. 2007). The study revealed that the overall physical activity of obese young adults has a mean of μ = 320 cpm μ = 320 cpm and a standard deviation of σ = 100 c p m . σ = 100 c p m . (In comparison, the mean for young adults of normal weight is 540 cpm.) In a random sample of n = 100 n = 100 obese young adults, consider the sample mean counts per minute, ¯ x x ‾ . Describe the sampling distribution of ¯ x x ‾ . What is the probability that the mean overall physical activity level of the sample is between 300 and 310 cpm? What is the probability that the mean overall physical activity level of the sample is greater than 360 cpm?
In: Math
Please Double Check answers I've recived 3 wrong answers on three diffrent questions today thank you
CNNBC recently reported that the mean annual cost of auto insurance is 1006 dollars. Assume the standard deviation is 245 dollars. You take a simple random sample of 73 auto insurance policies.
Find the probability that a single randomly selected value is less than 973 dollars. P(X < 973) =
Find the probability that a sample of size n = 73 is randomly selected with a mean less than 973 dollars. P(M < 973) =
In: Math
A sample containing years to maturity and yield for 40 corporate bonds are contained in the data given below.
Years to Maturity | Yield | Years to Maturity | Yield | |||
---|---|---|---|---|---|---|
23.50 | 4.757 | 3.75 | 2.769 | |||
21.75 | 2.473 | 12.00 | 6.293 | |||
21.50 | 4.464 | 17.50 | 7.411 | |||
23.50 | 4.684 | 18.00 | 3.558 | |||
27.00 | 4.799 | 8.25 | 0.945 | |||
18.25 | 3.755 | 23.25 | 2.966 | |||
15.75 | 7.068 | 14.75 | 1.476 | |||
2.00 | 7.043 | 10.00 | 1.382 | |||
8.75 | 6.540 | 23.00 | 6.334 | |||
5.25 | 7.000 | 15.25 | 0.887 | |||
11.25 | 4.823 | 4.75 | 4.810 | |||
25.75 | 1.874 | 18.00 | 1.238 | |||
14.25 | 5.654 | 3.00 | 6.767 | |||
19.25 | 1.745 | 9.50 | 3.745 | |||
25.00 | 8.153 | 17.50 | 4.186 | |||
6.75 | 6.571 | 17.00 | 5.991 | |||
23.00 | 7.506 | 9.50 | 7.322 | |||
19.00 | 2.857 | 5.50 | 4.871 | |||
10.75 | 8.010 | 27.50 | 2.403 | |||
21.25 | 4.214 | 26.00 | 4.500 |
a. What is the sample mean years to maturity for corporate bonds and what is the sample standard deviation?
Mean | ? (to 4 decimals) |
Standard deviation | ? (to 4 decimals) |
b. Develop a 95% confidence interval for the population mean years to maturity. Round the answer to four decimal places.
( , ) years
c. What is the sample mean yield on corporate bonds and what is the sample standard deviation?
Mean | ?(to 4 decimals) |
Standard deviation | ?(to 4 decimals) |
d. Develop a 95% confidence interval for the population mean yield on corporate bonds. Round the answer to four decimal places.
( , )percent
In: Math
As a data scientist of a company, you want to analyze the following data collected by your company which relates the advertising expenditure A in thousands of dollars to total sales S in thousands of dollars. The following table shows this relationship
Advertising Expenditure (A) | Total Sales (S) |
18.6 | 312 |
18.8 | 322 |
18.8 | 333 |
18.8 | 317 |
19 | 301 |
19 | 320 |
19.2 | 305 |
Using Advertising expenditure (A) as the domain and Total Sales (S) as the range, the data is not a function because the value 18.8 and 19 appear in the domain more than once with a different corresponding value of the range each time.
--Interpret the slope and y-intercept of this equation.
--Express this equation as a function S of A and find its domain.
--Predict the sales if the advertising expenditure is $25000.
In: Math
A prominent university conducted a survey on the effect of part-time work on student grade point average (GPA). Let x be the hours worked per week and y the GPA for the year. A summary of the results is below. What can the university conclude with an α of 0.05?
n = 21
sigmay = 55
,sigma x = 520
sigmay2 = 171
, aigmax2 = 15288
sigmayx = 1275
, sigma ( y − ŷ2) = 24
a) Compute the quantities below.
Bhat0 = , Bhat =
What GPA is predicted when a students works 9 hours a week?
b) Compute the appropriate test statistic(s) for H1: β < 0.
Critical value = ; Test statistic =
Decision:
---Select---
Reject H0
Fail to reject H0
c) Compute the corresponding effect size(s) and indicate magnitude(s).
If not appropriate, input and/or select "na" below.
Effect size = ;
---Select---
na
trivial effect
small effect
medium effect
large effect
d) Make an interpretation based on the results.
More hours of part-time work significantly predicts a higher GPA.
More hours of part-time work significantly predicts a lower GPA.
Part-time work does not significantly predict GPA.
In: Math
Financial analysts know that January credit card charges will generally be much lower than those of the month before. What about the difference between January and the next month? Does the trend continue? The accompanying data set contains the monthly credit card charges of a random sample of 99 cardholders. Complete parts a) through e) below.
January |
February |
|
---|---|---|
902.74 |
641.04 |
|
7212.18 |
4565.35 |
|
4235.42 |
2270.56 |
|
79.92 |
300.09 |
|
4045.57 |
1377.72 |
|
89.29 |
−120.74 |
|
3289.59 |
1928.85 |
|
2419.54 |
2609.97 |
|
83.81 |
144.83 |
|
6.42 |
392.85 |
|
0.00 |
40.46 |
|
564.69 |
295.63 |
|
2712.23 |
848.62 |
|
187.12 |
162.12 |
|
3265.86 |
2412.45 |
|
1523.59 |
956.31 |
|
1359.23 |
38.03 |
|
733.33 |
2656.79 |
|
75.09 |
64.94 |
|
70.29 |
−70.32 |
|
634.53 |
1862.61 |
|
1041.23 |
478.07 |
|
553.08 |
994.64 |
|
1016.27 |
774.54 |
|
1304.94 |
3368.08 |
|
249.39 |
5.52 |
|
48.78 |
96.93 |
|
872.34 |
890.89 |
|
485.94 |
485.21 |
|
616.52 |
1485.52 |
|
1574.18 |
890.46 |
|
422.34 |
391.43 |
|
770.85 |
323.19 |
|
56.53 |
0.00 |
|
1486.78 |
2253.73 |
|
495.28 |
390.19 |
|
1064.88 |
1065.85 |
|
510.65 |
131.33 |
|
5637.68 |
4942.63 |
|
5.49 |
5.51 |
|
871.63 |
591.44 |
|
1636.66 |
3364.19 |
|
92.13 |
85.99 |
|
669.34 |
1367.13 |
|
829.32 |
280.85 |
|
69.24 |
67.99 |
|
830.54 |
1057.56 |
|
2301.44 |
3317.76 |
|
270.67 |
14.13 |
|
210.42 |
160.52 |
|
1012.36 |
519.35 |
|
1044.96 |
2021.35 |
|
298.64 |
635.44 |
|
−29.99 |
0.00 |
|
1634.61 |
393.34 |
|
1731.93 |
1323.33 |
|
0.00 |
65.16 |
|
31.43 |
28.75 |
|
4.95 |
77.15 |
|
1088.69 |
892.78 |
|
26.88 |
29.03 |
|
120.31120.31 |
32.23 |
|
2007.48 |
815.63 |
|
291.31 |
779.47 |
|
104.02 |
0.00 |
|
53.01 |
66.25 |
|
2842.52 |
1530.91 |
|
675.47 |
293.45 |
|
221.86 |
171.92 |
|
37.79 |
4.78 |
|
533.25 |
880.96 |
|
1932.71 |
1063.55 |
|
692.17 |
915.55 |
|
6804.35 |
5941.41 |
|
393.36 |
466.47 |
|
1309.18 |
302.89 |
|
796.21 |
497.02 |
|
0.00 |
266.64 |
|
1040.29 |
59.45 |
|
565.12 |
206.62 |
|
339.14 |
412.34 |
|
5275.34 |
5324.54 |
|
40.09 |
72.58 |
|
43.39 |
38.45 |
|
653.63 |
480.25 |
|
1071.23 |
416.29 |
|
2337.04 |
1787.19 |
|
91.47 |
175.32 |
|
1433.01 |
1107.78 |
|
719.86 |
307.79 |
|
28.61 |
24.19 |
|
980.34 |
1216.35 |
|
1576.18 |
1810.23 |
|
0.00 |
468.24 |
|
161.96 |
147.68 |
|
494.32 |
1995.28 |
|
534.11 |
935.24 |
|
462.45 |
114.51 |
|
1478.23 |
2093.37 |
a) Build a regression model to predict February charges from January charges.
Feb=____+____Jan (Round to two decimal places as needed.)
Check the conditions for this model. Select all of the true statements related to checking the conditions.
A. All of the conditions are definitely satisfied.
B. The Linearity Condition is not satisfied.
C. The Randomization Condition is not satisfied.
D. The Equal Spread Condition is not satisfied.
E. The Nearly Normal Condition is not satisfied.
b) How much, on average, will cardholders who charged $2000 in January charge in February? $____ (Round to the nearest cent as needed.)
c) Give a 95% confidence interval for the average February charges of cardholders who charged $2000 in January.
($___,$___) (Round to the nearest cent as needed.)
d) From part c), give a 95% confidence interval for the average decrease in the charges of cardholders who charged $2000 in January.
($___,$___) (Round to the nearest cent as needed.)
e) What reservations, if any, would a researcher have about the confidence intervals made in parts c) and d)? Select all that apply.
A. The residuals show increasing spread, so the confidence intervals may not be valid.
B. The residuals show a curvilinear pattern, so the confidence intervals may not be valid.
C. The data are not linear, so the confidence intervals are not valid.
D. The data are not independent, so the confidence intervals are not valid.
E. A researcher would not have any reservations. The confidence intervals are valid. Click to select your answer(s).
In: Math