The Florida Company was flooded by a Hurricane and somehow lost
part of their forecasting data. The missing values in the following
table must be recalculated from the remaining data. Exponential
smoothing constant of 0.35 is used for both the forecast and
updated MAD.
NOTE: All forecast, error, MAD, and TS values
should be rounded to the nearest hundredth (two decimal after dot,
for example 99.99)
Period |
Actual |
Forecast |
Error |
MADt |
TSt |
0 |
9.00 |
||||
1 |
260.00 |
255.00 |
5.00 |
0.66 |
|
2 |
270.00 |
256.75 |
13.25 |
9.58 |
|
3 |
260.00 |
Based on -/+ 4.00 action limits, is the forecasting process under
control? (Enter “YES” or “NO”)
In: Statistics and Probability
1. A research institute poll asked respondents if they felt vulnerable to identity theft. In the poll of 968 randomly selected American adults, 511 of them responded with “yes.” a). Find the 95% confidence interval for the proportion of adults who felt vulnerable to identify theft .
b). Write a statement that correctly interprets the confidence interval. Choose the correct answer below.
A. One has 95% confidence that the sample proportion is equal to the population proportion.
B. One has 95% confidence that the interval from the lower bound to the upper bound actually does contain the true value of the population proportion.
In: Statistics and Probability
A survey of 20 randomly sampled judges employed by the state of Florida found that they earned an average wage (including benefits) of $63.00 per hour. The sample standard deviation was $5.90 per hour. (Use t Distribution Table.)
What is the best estimate of the population mean?
Develop a 98% confidence interval for the population mean wage (including benefits) for these employees. (Round your answers to 2 decimal places.)
Confidence interval for the population mean wage is between and
How large a sample is needed to assess the population mean with an allowable error of $2.00 at 95% confidence? (Round up your answer to the next whole number.)
Sample Size
In: Statistics and Probability
The StatCrunch data set for this question contains the data
measurements described in Question 11. (H0 :
µ1 - µ2 ≤ 0
HA : µ1 - µ2 > 0)
Assume that the two samples are dawn from independent, normally
distributed populations that have different standard deviations.
Use this data set and the results from Question 11 to calculate the
p-value for the hypothesis test. Round your answer to three
decimal places; add trailing zeros as needed.
The p-value = [S90PValue].
Using the result from Question 12, identify the appropriate decision for the hypothesis test in Question 11, along with its interpretation. Use α = 0.01.
Fail to reject H0. There is insufficient evidence to support the original claim that the mean amount of Strontium-90 collected from City 1 residents' teeth is greater than the mean amount collected from the residents of City 2.
Reject H0. There is insufficient evidence to support the original claim that the mean amount of Strontium-90 collected from City 1 residents' teeth is greater than the mean amount collected from the residents of City 2.
Fail to reject H0. There is insufficient evidence to reject the original claim that the mean amount of Strontium-90 collected from City 1 residents' teeth is greater than the mean amount collected from the residents of City 2.
Reject H0. There is sufficient evidence to support the original claim that the mean amount of Strontium-90 collected from City 1 residents' teeth is greater than the mean amount collected from the residents of City 2.
city1 city 2
104 117 ""
86 73 ""
121 100 ""
119 85 ""
101 84 ""
104 107 ""
213 110 ""
144 111 ""
290 105 ""
100 133 ""
275 101 ""
145 209 ""
In: Statistics and Probability
According to the census bureau publication: current construction reports, the mean price of new mobile homes is 63,500. The standard deviation of the prices is 5,500.
a). For samples 100 new mobile homes, determine the mean and standard deviation of all possible sample mean prices (hint: find u-x and o-x)
b). Repeat part a for samples of size 200
In: Statistics and Probability
Data from the past shows that on average, a ready-mixed concrete plant receives 100 orders for concrete every year. The maximum number of orders that the plant can fulfil each week is 2. (a) What is the probability that in a given week the plant cannot fulfil all the placed orders? (b) Assume the answer to part (a) is 20% (It is not; I just want to make sure that everybody uses the same number for part (b)). Suppose there are 5 of such plants. What is the probability that in a given week 2 of the plants cannot fulfill their orders
In: Statistics and Probability
SALARY | EDUC | EXPER | TIME |
39000 | 12 | 0 | 1 |
40200 | 10 | 44 | 7 |
42900 | 12 | 5 | 30 |
43800 | 8 | 6 | 7 |
43800 | 8 | 8 | 6 |
43800 | 12 | 0 | 7 |
43800 | 12 | 0 | 10 |
43800 | 12 | 5 | 6 |
44400 | 15 | 75 | 2 |
45000 | 8 | 52 | 3 |
45000 | 12 | 8 | 19 |
46200 | 12 | 52 | 3 |
48000 | 8 | 70 | 20 |
48000 | 12 | 6 | 23 |
48000 | 12 | 11 | 12 |
48000 | 12 | 11 | 17 |
48000 | 12 | 63 | 22 |
48000 | 12 | 144 | 24 |
48000 | 12 | 163 | 12 |
48000 | 12 | 228 | 26 |
48000 | 12 | 381 | 1 |
48000 | 16 | 214 | 15 |
49800 | 8 | 318 | 25 |
51000 | 8 | 96 | 33 |
51000 | 12 | 36 | 15 |
51000 | 12 | 59 | 14 |
51000 | 15 | 115 | 1 |
51000 | 15 | 165 | 4 |
51000 | 16 | 123 | 12 |
51600 | 12 | 18 | 12 |
52200 | 8 | 102 | 29 |
52200 | 12 | 127 | 29 |
52800 | 8 | 90 | 11 |
52800 | 8 | 190 | 1 |
52800 | 12 | 107 | 11 |
54000 | 8 | 173 | 34 |
54000 | 8 | 228 | 33 |
54000 | 12 | 26 | 11 |
54000 | 12 | 36 | 33 |
54000 | 12 | 38 | 22 |
54000 | 12 | 82 | 29 |
54000 | 12 | 169 | 27 |
54000 | 12 | 244 | 1 |
54000 | 15 | 24 | 13 |
54000 | 15 | 49 | 27 |
54000 | 15 | 51 | 21 |
54000 | 15 | 122 | 33 |
55200 | 12 | 97 | 17 |
55200 | 12 | 196 | 32 |
55800 | 12 | 133 | 30 |
56400 | 12 | 55 | 9 |
57000 | 12 | 90 | 23 |
57000 | 12 | 117 | 25 |
57000 | 15 | 51 | 17 |
57000 | 15 | 61 | 11 |
57000 | 15 | 241 | 34 |
60000 | 12 | 121 | 30 |
60000 | 15 | 79 | 13 |
61200 | 12 | 209 | 21 |
63000 | 12 | 87 | 33 |
63000 | 15 | 231 | 15 |
46200 | 12 | 12 | 22 |
50400 | 15 | 14 | 3 |
51000 | 12 | 180 | 15 |
51000 | 12 | 315 | 2 |
52200 | 12 | 29 | 14 |
54000 | 12 | 7 | 21 |
54000 | 12 | 38 | 11 |
54000 | 12 | 113 | 3 |
54000 | 15 | 18 | 8 |
54000 | 15 | 359 | 11 |
57000 | 15 | 36 | 5 |
60000 | 8 | 320 | 21 |
60000 | 12 | 24 | 2 |
60000 | 12 | 32 | 17 |
60000 | 12 | 49 | 8 |
60000 | 12 | 56 | 33 |
60000 | 12 | 252 | 11 |
60000 | 12 | 272 | 19 |
60000 | 15 | 25 | 13 |
60000 | 15 | 36 | 32 |
60000 | 15 | 56 | 12 |
60000 | 15 | 64 | 33 |
60000 | 15 | 108 | 16 |
60000 | 16 | 46 | 3 |
63000 | 15 | 72 | 17 |
66000 | 15 | 64 | 16 |
66000 | 15 | 84 | 33 |
66000 | 15 | 216 | 16 |
68400 | 15 | 42 | 7 |
69000 | 12 | 175 | 10 |
69000 | 15 | 132 | 24 |
81000 | 16 | 55 |
SUMMARY OUTPUT | ||||||||
Regression Statistics | ||||||||
Multiple R | 0.41198516 | |||||||
R Square | 0.16973178 | |||||||
Adjusted R Square | 0.16060795 | |||||||
Standard Error | 6501.12045 | |||||||
Observations | 93 | |||||||
ANOVA | ||||||||
df | SS | MS | F | Significance F | ||||
Regression | 1 | 786253429 | 786253429 | 18.60313 | 4.08E-05 | |||
Residual | 91 | 3.85E+09 | 42264567.1 | |||||
Total | 92 | 4.63E+09 | ||||||
Coefficients | Standard Error | t Stat | P-value | Lower 95% | Upper 95% | Lower 95.0% | Upper 95.0% | |
Intercept | 38185.5979 | 3774.3766 | 10.117061 | 1.45E-16 | 30688.26252 | 45682.93 | 30688.26 | 45682.93 |
X Variable 1 | 1280.85932 | 296.96712 | 4.31313512 | 4.08E-05 | 690.9706164 | 1870.748 | 690.9706 | 1870.748 |
This data set was obtained by collecting information on a randomly selected sample of 93 employees working at a bank.
SALARY- starting annual salary at the time of hire
EDUC - number of years of schooling at the time of the hire
EXPER - number of months of previous work experience at the time of hire
TIME - number of months that the employee has been working at the bank until now
2. Use the least squares method to fit a simple linear model that relates the salary (dependent variable) to education (independent variable).
a- What is your model? State the hypothesis that is to be tested, the decision rule, the test statistic, and your decision, using a level of significance of 5%.
b – What percentage of the variation in salary has been explained by the regression?
c – Provide a 95% confidence interval estimate for the true slope value.
d - Based on your model, what is the expected salary of a new hire with 12 years of education?
e – What is the 95% prediction interval for the salary of a new hire with 12 years of education? Use the fact that the distance value = 0.011286
In: Statistics and Probability
The art of institutions of the development of the world.
1. The smaller, or more restricted, a range of scores, the more likely that the Pearson correlation will overestimate the true strength of the relationship between two variables.. True or False
2. 2 scholars are studying a sample of 50 recently discovered
aliens quantify their aggression on a scale of 1-100, with 100
being extremely aggressive. With the mean of 65 and STD of 6, range
of scores that fall between the z-scores of -1.50 and +1.50 is 24.
True or False
3. A regression line indicates the predicted Y score for each X score. True or False
4.Which relationship is stronger, r = +0.62 or r = –0.62?
In: Statistics and Probability
Stanley’s Secretarial Services uses an average of 1,215 boxes of envelopes each year. Carrying costs per box are $75 per year, and it costs $10 to place and receive an order. |
a. |
What is the optimal order quantity for boxes of envelopes, to the nearest whole number? |
Economic order quantity | boxes |
b. |
What is the average number of boxes of envelopes in inventory, to the nearest whole number? |
Average number of boxes |
c. |
How times will Stanley order envelopes each year (to the nearest whole number)? |
Number of orders per year |
d. |
Compute the total of ordering and carrying costs for boxes of envelopes. (Round to the nearest whole number, but do not type the dollar sign.) |
Total cost | $ |
e. |
Calculate the change in total annual cost, if per-box holding costs went up by $9 per year. (Round your intermediate calculations to 2 decimal places and final answer to 2 decimal places, but do not type the dollar sign.) |
Increase by | $ |
In: Statistics and Probability
Two plots at Rothamsted Experimental Station were studied for production of wheat straw. For a random sample of years, the annual wheat straw production (in pounds) from one plot was as follows. 6.40 7.03 6.47 5.70 7.31 7.18 7.06 5.79 6.24 5.91 6.14 Use a calculator to verify that, for this plot, the sample variance is s2 ≈ 0.341. Another random sample of years for a second plot gave the following annual wheat production (in pounds). 6.40 7.10 5.91 5.84 7.22 5.58 5.47 5.86 Use a calculator to verify that the sample variance for this plot is s2 ≈ 0.447. Test the claim that there is a difference (either way) in the population variance of wheat straw production for these two plots. Use a 5% level of signifcance. (a) What is the level of significance? State the null and alternate hypotheses. Ho: σ12 = σ22; H1: σ12 > σ22 Ho: σ12 > σ22; H1: σ12 = σ22 Ho: σ22 = σ12; H1: σ22 > σ12 Ho: σ12 = σ22; H1: σ12 ≠ σ22 (b) Find the value of the sample F statistic. (Use 2 decimal places.) What are the degrees of freedom? dfN dfD What assumptions are you making about the original distribution? The populations follow independent normal distributions. The populations follow independent normal distributions. We have random samples from each population. The populations follow dependent normal distributions. We have random samples from each population. The populations follow independent chi-square distributions. We have random samples from each population. (c) Find or estimate the P-value of the sample test statistic. (Use 4 decimal places.) p-value > 0.200 0.100 < p-value < 0.200 0.050 < p-value < 0.100 0.020 < p-value < 0.050 0.002 < p-value < 0.020 p-value < 0.002 (d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis? At the α = 0.05 level, we reject the null hypothesis and conclude the data are not statistically significant. At the α = 0.05 level, we reject the null hypothesis and conclude the data are statistically significant. At the α = 0.05 level, we fail to reject the null hypothesis and conclude the data are not statistically significant. At the α = 0.05 level, we fail to reject the null hypothesis and conclude the data are statistically significant. (e) Interpret your conclusion in the context of the application. Fail to reject the null hypothesis, there is sufficient evidence that the variance in annual wheat production differs between the two plots. Reject the null hypothesis, there is insufficient evidence that the variance in annual wheat production differs between the two plots. Reject the null hypothesis, there is sufficient evidence that the variance in annual wheat production differs between the two plots. Fail to reject the null hypothesis, there is insufficient evidence that the variance in annual wheat production differs between the two plots.
In: Statistics and Probability
An airline estimates that 80% of passengers who reserve the tickets actually show up for the flights. Based on this information, it has to decide how many tickets it will sell for each flight, which is typically more than the number of seats actually available. In the economy section of a particular aircraft, 200 seats are available. The airline sells 225 seats. What is the probability that more passengers will show up than there are seats for?
In: Statistics and Probability
The quantitative reasoning GRE scores are known to approximately have a Normal distribution with a mean of = 151.3 points and a standard deviation of = 8.7 points.
a. Use the Empirical Rule to specify the ranges into which 68%, 95%, and 99.7% of test takers fall. Include a picture to illustrate the ranges.
b. A graduate program in Public Policy Analysis admits only students with quantitative reasoning GRE scores in the top 30%. What is the lowest GRE score the program will accept?
c. Above what score do the top 1% of GRE scores fall?
In: Statistics and Probability
3,500 women between the ages 60 -74 years are in a town consisting of a population of 15,000 persons. 85 cases of cancer are in the town one year and 30 of these cases were women 60 -74 years old. What is the prevalence of cancer among women of this age group?
In: Statistics and Probability
Since an instant replay system for tennis was introduced at a major tournament, men challenged 1441 referee calls, with the result that 422 of the calls were overturned. Women challenged 758 referee calls, and 216 of the calls were overturned. Use a 0.05 significance level to test the claim that men and women have equal success in challenging calls. Complete parts (a) through (c) below.
a. Test the claim using a hypothesis test.
Consider the first sample to be the sample of male tennis players who challenged referee calls and the second sample to be the sample of female tennis players who challenged referee calls. What are the null and alternative hypotheses for the hypothesis test?
Identify the test statistic.
Identify the P-value.
What is the conclusion based on the hypothesis test?
The P-value is
(less than or greater than)
the significance level of
alpha =0.05, so
(reject OR fail to reject) the null hypothesis. There
(is sufficient OR is not sufficient)
evidence to warrant rejection of the claim that women and men have equal success in challenging calls.
b. Test the claim by constructing an appropriate confidence interval.
The
9595%
confidence interval is
nothingless than<left parenthesis p 1 minus p 2 right parenthesisp1−p2less than<nothing.
(Round to three decimal places as needed.)
What is the conclusion based on the confidence interval?
Because the confidence interval limits
(include OR do not include)
0, there
(does not OR does)
appear to be a significant difference between the two proportions. There
(is not sufficient OR is sufficient)
evidence to warrant rejection of the claim that men and women have equal success in challenging calls.
c. Based on the results, does it appear that men and women may have equal success in challenging calls?
A.
The confidence interval suggests that there is no significant difference between the success of men and women in challenging calls.
B.
The confidence interval suggests that there is a significant difference between the success of men and women in challenging calls. It is reasonable to speculate that women have more success.
C.
The confidence interval suggests that there is a significant difference between the success of men and women in challenging calls. It is reasonable to speculate that men have more success.
D.
There is not enough information to reach a conclusion.
In: Statistics and Probability
Determine the critical value.
Using the z-tables (or t-tables), determine the critical value for the left-tailed z-test with α=0.01
Question options:
a)-2.326 |
|
b)2.326 |
|
c)2.576 |
|
d)-2.576 |
In: Statistics and Probability