A certain retail store bases its staffing on the number of customers that arrive during certain time slots. Based on prior experience this store expects 32% of its customers between 8:00 am and 12:00 pm; 21% of its customers between 12:00 pm and 4:00 pm; 35% of its customers between 4:00 pm and 8:00 pm; and 12% of its customers between 8:00 pm and midnight. On a certain day, the store had 214, 198, 276, and 134 customers in those time slots, respectively. Should the store change its staffing? (Consider an alpha of 0.05.)
Solution:
Ho: The expected values match the observed values
Ha: The expected values do not match the observed values
assign(“exp”,c(32,21,35,12)) assign(“obs”,c(214,198,276,134))
sum((obs-exp)^2/exp) = 5426.773
1-pchisq(5426.773,3) = 0
p < alpha, therefore RHo: the store should change its
staffing.
What was wrong with this solution?
In: Statistics and Probability
the following are the monthly sales (in thousands of dollars) for a company in four regions of the country.
|
North |
East |
South |
West |
|
34 28 18 24 |
47 36 30 38 44 |
40 30 41 29 |
21 30 24 37 23 |
Does the data suggest that there is a difference in the mean monthly sales among the different regions?
In: Statistics and Probability
1)
Identify the class width for the given Frequency Distribution Table
List -------------Frequency
10 to 14 -------------27
15 to 19 -------------35
20 to 24 -------------14
25 to 29 --------------2
30 to 34 --------------4
35 to 39 --------------2
40 to 44 --------------1
2)
Which measurement is the most accurate?
Mode
Mean
Range
Median
In: Statistics and Probability
Dorothy Kelly sells life insurance for the Prudence Insurance Company. She sells insurance by making visits to her clients homes. Dorothy believes that the number of sales should depend, to some degree, on the number of visits made. For the past several years, she kept careful records of the number of visits (x) she made each week and the number of people (y) who bought insurance that week. For a random sample of 15 such weeks, the x and y values follow.
| x | 11 | 17 | 17 | 14 | 28 | 5 | 20 | 14 | 22 | 7 | 15 | 29 | 8 | 25 | 16 |
| y | 2 | 13 | 9 | 3 | 8 | 2 | 5 | 6 | 8 | 3 | 5 | 10 | 6 | 10 | 7 |
Σx = 248; Σy = 97; Σx2 = 4,844; Σy2 = 775; Σxy = 1,828
(a) Find x, y, b, and the equation of the least-squares line. (Round your answers for x and y to two decimal places. Round your least-squares estimates to three decimal places.)
| x | = | |
| y | = | |
| b | = | |
| ŷ | = | + x |
(b) Draw a scatter diagram for the data. Plot the least-squares
line on your scatter diagram.
(c) Find the sample correlation coefficient r and the
coefficient of determination. (Round your answers to three decimal
places.)
| r = | |
| r2 = |
What percentage of variation in y is explained by the
least-squares model? (Round your answer to one decimal
place.)
%
(d) In a week during which Dorothy makes 21 visits, how many people
do you predict will buy insurance from her? (Round your answer to
one decimal place.)
people
In: Statistics and Probability
5.90 Genetics of peas. According to genetic theory, the blossom color in the second generation of a certain cross of sweet peas should be red or white in a 3:1 ratio. That is, each plant has probability 3/4 of having red blossoms, and the blossom colors of separate plants are independent. (a) What is the probability that exactly 8 out of 10 of these plants have red blossoms? (b) What is the mean number of red-blossomed plants when 130 plants of this type are grown from seeds? (c) What is the probability of obtaining at least 90 red-blossomed plants when 130 plants are grown from seeds?
In: Statistics and Probability
A multiple regression model is to be constructed to predict the final exam score of a university student doing a particular course based upon their mid-term exam score, the average number of hours spent studying per week and the average number of hours spent watching television per week.
Data has been collected on 30 randomly selected individuals: hide data
Download the data
| Final score | Mid-term Score | Hours studying per week |
Hours watching TV per week |
|---|---|---|---|
| 76 | 85 | 19 | 34 |
| 60 | 85 | 3 | 11 |
| 42 | 63 | 10 | 33 |
| 32 | 40 | 6 | 12 |
| 46 | 65 | 6 | 16 |
| 48 | 72 | 13 | 30 |
| 30 | 37 | 14 | 33 |
| 47 | 47 | 9 | 26 |
| 33 | 26 | 19 | 7 |
| 60 | 65 | 18 | 8 |
| 59 | 79 | 13 | 24 |
| 28 | 29 | 10 | 22 |
| 24 | 33 | 7 | 9 |
| 59 | 77 | 5 | 6 |
| 66 | 93 | 17 | 7 |
| 51 | 48 | 18 | 29 |
| 74 | 98 | 4 | 7 |
| 29 | 23 | 8 | 15 |
| 31 | 31 | 6 | 9 |
| 69 | 79 | 18 | 15 |
| 60 | 73 | 3 | 6 |
| 62 | 89 | 11 | 31 |
| 49 | 52 | 19 | 6 |
| 37 | 44 | 14 | 9 |
| 63 | 94 | 10 | 26 |
| 62 | 89 | 7 | 27 |
| 30 | 31 | 18 | 32 |
| 42 | 60 | 14 | 17 |
| 54 | 70 | 4 | 22 |
| 73 | 97 | 19 | 28 |
a)Find the multiple regression equation using all three explanatory variables. Assume that X1 is mid-term score, X2 is hours studying per week and X3 is hours watching television per week. Give your answers to 3 decimal places.
y^ = + mid-term score + hours studying + hours watching television
b)At a level of significance of 0.05, the result of the F test for this model is that the null hypothesis isis not rejected.
For parts c) and d), using the data, separately calculate the correlations between the response variable and each of the three explanatory variables.
c)The explanatory variable that is most correlated with final score is:
mid-term score
hours studying per week
hours watching television per week
d)The explanatory variable that is least correlated with final score is:
mid-term score
hours studying per week
hours watching television per week
e)The value of R2 for this model, to 2 decimal places, is equal to
f)The value of se for this model, to 3 decimal places, is equal to
g)Construct a new multiple regression model by removing the variable average hours spent watching television per week. Give your answers to 3 decimal places.
The new regression model equation is:
y^ = + mid-term score + hours studying
h)In the new model compared to the previous one, the value of R2 (to 2 decimal places) is:
increased
decreased
unchanged
i)In the new model compared to the previous one, the value of se (to 3 decimal places) is:
increased
decreased
unchanged
In: Statistics and Probability
Calculate the present value of the following amounts assuming that the cash flow is received at the end of each specified period: 211
| Situation | Cash Flow | Discount rate | End of period( pear year) | Frequency |
| A | $15,000 | 10% | 8 | Bimonthly |
| B | $20,000 | 13% | 12 | Semiannual |
| C | $6,000 | 9% | 10 | Quarterly |
| D | $50,000 | 7% | 50 | Annual |
| E | $8,500 | 7% | 15 | Quarterly |
I appreciate if you can explain me step by step. Thank you.
In: Finance
Write a one-page essay on customer expectations. What are customers’ expectations when they walk into these places of business?
Choose any two:
Furniture store
Restaurant
Medical clinic
Pharmacy
Convenience store (e.g. 7-Eleven)
Reference your own experience and material from the text. Remember that spelling and grammar are important.
In: Accounting
Some five courses are offered on campus in a semester. The group of students who take at least one of these courses consists of 155 students. There are 80 students registered in each course. For any two among these courses, there are precisely 40 students who take both of them. For any three of these courses, there are precisely 20 students who take all three of them. For any four among these courses, there are precisely 10 students who take every one of these four courses. How many students in this group take every one among the five courses in this memorable semester?
In: Statistics and Probability
Identifying individuals with a high risk of Alzheimer’s disease usually involves a long series of cognitive tests. However, researchers have developed a 7-Minute Screen, which is a quick and easy way to accomplish the same goal. The question is whether the 7-Minute Screen is as effective as the complete series of tests. To address this question, Ijuin et al. (2008) administered both tests to a group of patients and compared the results. The following data represent results similar to those obtained in the study.
|
Patient |
7-Minute Screen |
Cognitive Series |
|
|
A |
3 |
11 |
|
|
B |
8 |
19 |
|
|
C |
10 |
22 |
|
|
D |
8 |
20 |
|
|
E |
4 |
14 |
|
|
F |
7 |
13 |
|
|
G |
4 |
9 |
|
|
H |
5 |
20 |
|
|
I |
14 |
25 |
|
A. Compute the Pearson Correlation Coefficient for these data.
r =
B. The correlation is
Medium and negative
Large and negative
Large and positive
Medium and positive
C. The null hypothesis for a two-tailed test is
The correlation between the scores on the 7-minute screen test and the scores on the cognitive series is significant
The correlation between the scores on the 7-minute screen test and the scores on the cognitive series is not significant
The correlation between the scores on the 7-minute screen test and the scores on the cognitive series is less than zero
The correlation between the scores on the 7-minute screen test and the scores on the cognitive series is different from zero
D. If we use a two-tailed test with α = .01, the critical r values are:
±0.666
±0.582
±0.798
±0.750
E. Your decision is:
Reject the null hypothesis and conclude that there is a significant correlation between the scores on the 7-minute screen test and the scores on the cognitive series
Fail to reject the null hypothesis and conclude that there is a significant correlation between the scores on the 7-minute screen test and the scores on the cognitive series
Reject the null hypothesis and conclude that there is no significant correlation between the scores on the 7-minute screen test and the scores on the cognitive series
Fail to reject the null hypothesis and conclude that there is no significant correlation between the scores on the 7-minute screen test and the scores on the cognitive series
In: Statistics and Probability