|
The following data are from a completely randomized design.
a. Compute the sum of squares between treatments. b. Compute the mean square between treatments. c. Compute the sum of squares due to error. d. Compute the mean square due to error (to 1 decimal). e. Set up the ANOVA table for this problem. Round all Sum of Squares to nearest whole numbers. Round all Mean Squares to one decimal place. Round F to two decimal places. Round your p-value to 4 decimal places.
f. At the level of significance, test whether the means for the three treatments are equal. Calculate the value of the test statistic (to 2 decimals). The -value is - Select your answer -less than .01between .01 and .025between .025 and .05between .05 and .10greater than .10Item 16 What is your conclusion? - Select your answer -Conclude that not all treatment means are equalDo not reject the assumption that the means for all three treatments are equalItem 17 |
||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
In: Statistics and Probability
In a study of color perception, 356 men are tested, and 39 are found to have red/green color blindness.
1. n ˆ p ( 1 − ˆ p ) =
2. ˆ p =
3. Is n ˆ p ( 1 − ˆ p ) ≥ 10 ?( yes or no)
4. The margin of error is use the 85 % confidence level
5.Construct a 85 % confidence interval for the percent of men in general population who are color blind. Lower Bound Upper Bound
In: Statistics and Probability
Consider the experimental results for the following randomized block design. Make the calculations necessary to set up the analysis of variance table.
| Treatment | ||||
| A | B | C | ||
| 1 | 11 | 10 | 8 | |
| 2 | 12 | 6 | 6 | |
| Blocks | 3 | 18 | 16 | 15 |
| 4 | 20 | 18 | 18 | |
| 5 | 9 | 7 | 9 | |
Use a=.05 to test for any significant differences. Show entries to 2 decimals, if necessary. Round all intermediate values to two decimal places in your calculations. If your answer is zero enter "0".
| Source of Variation | Sum of Squares | Degrees of Freedom | Mean Square | F | p-value |
| Treatments | |||||
| Blocks | |||||
| Error | |||||
| Total |
The -value is - Select your answer -less than .01between .01 and .025between .025 and .05between .05 and .10greater than .10Item 14
What is your conclusion?
- Select your answer - Conclude not all treatment means are
equal Do not reject the assumption that the treatment
means are equal Item 15
In: Statistics and Probability
According to the Air Transport Association of America, the
average operating cost of an MD-80 jet airliner is $2,087 per hour.
Suppose the operating costs of an MD-80 jet airliner are normally
distributed with a standard deviation of $169 per hour.
(Round the value of z to 2 decimal places. Round your
answers to 2 decimal places.)
| (a) At what
operating cost would only 20% of the operating costs be less? $enter the dollar amount at which only 20% of the operating costs would be less (b) At what operating cost would 65% of the operating costs be more? $enter the dollar amount at which 65% of the operating costs would be more (c) What operating cost would be more than 85% of operating costs? $enter the dollar amount that would be more than 85% of operating costs |
In: Statistics and Probability
Consider the data set.
(a)
Find the range. (Enter an exact number.)
(b)
Use the defining formula to compute the sample standard
deviation s. (Enter a number. Round your answer to two
decimal places.)
(c)
Use the defining formula to compute the population standard deviation σ. (Enter a number. Round your answer to two decimal places.)
In: Statistics and Probability
The mean caloric intake of an adult male is 2800 with a standard deviation of 115. To verify this information, a sample of 25 men are selected and determined to have a mean caloric intake of 2950. Determine the 98% confidence interval for mean caloric intake of an adult male.
Solution:
Since n < 30, a t-test must be used.
2950 – qt(1.98/2,24)*115/sqrt(25)
2950 + qt(1.98/2,24)*115/sqrt(25)
[2892.68, 3007.32]
What is wrong with this solution?
In: Statistics and Probability
Suppose we are interested in whether there is a difference between the median numbers of hours spent each week by men and woman watching television. Two random samples were taken: the numbers of hours taken by men and women watching television are shown below. You are about to test the null hypothesis that there is no difference in weekly television watching hours between men and woman. Men: 5 10 12 15; Women: 0 7 4 3 6 8
a) calculate the rank sum for the smaller sample;
b) calculate the z statistic to test the null hypothesis; (I don't know how to get m for this question so do I use a different formula?)
c) what is the p value according to the above test?
d) give the possible range for the rank sum for the smaller sample
In: Statistics and Probability
The lengths of a particular animal's pregnancies are approximately normally distributed, with mean muequals257 days and standard deviation sigma equals20 days. (a) What proportion of pregnancies lasts more than 267 days? (b) What proportion of pregnancies lasts between 242 and 272 days? (c) What is the probability that a randomly selected pregnancy lasts no more than 217 days? (d) A "very preterm" baby is one whose gestation period is less than 227 days. Are very preterm babies unusual? (a) The proportion of pregnancies that last more than 267 days is nothing.
In: Statistics and Probability
Briefly discuss the implications of a type 1 error and of a type 2 error.
|
H0: The average ‘freshness’ of an edible food arrangement is at least 3 hours |
|
H1: The average ‘freshness’ of an edible food arrangement is not 3 hours |
|
H0: The average ‘freshness’ of an edible food arrangement is more than 3 hours |
|
H1: The average ‘freshness’ of an edible food arrangement is less than 3 hours |
|
H0: The average number of faulty watches per 1,000 watches is less than 4 |
|
H1: The average number of faulty watches per 1,000 watches is at least 4 |
|
H0: The average number of faulty watches per 1,000 watches is 4 |
|
H1: The average number of faulty watches per 1,000 watches is more than 4 |
In: Statistics and Probability
(10) 2. ESP (extra sensory perception) is the ability to read minds. We have a set of cards with each card having one of 5 shapes on them (square, rectangle, triangle, circle, question mark) and there are an equal number of each shape in the deck. We are going to select one card from the deck, stare at it and then there is a guy who claims he has ESP who is going to guess at the shape on the card we drew out (after each time we pull a card out – the ESP dude will make a guess- then we put the card back ). We did this experiment 250 times and we count how many times he guesses correctly
a) Why does this an experiment that fits the criterion of a binomial?
b) What is the mean, standard deviation and normal range for the number of times he guesses correctly
c) If he guesses 55 correct do you think he has ESP?
Explain
In: Statistics and Probability
Smartphone adoption among American younger adults has increased substantially and mobile access to the internet is pervasive. 17% of young adults, ages 18-29, who own a smartphone are “smartphone – dependent”, meaning that they do not have home broadband service and have limited options for going online other than their mobile device.
If a sample of 10 American young adults is selected what is the probability that
a. 3 are smartphone-dependent?
b. At least 3 are smartphone-dependent?
c. At most 6 are smartphone-dependent?
d. If you selected the sample in a particular geographical area and found that none of the 10 respondents are smartphone-dependent, what conclusions might you reach about whether the percentage of smartphone-dependent young adults in this area was 17%?
PLEASE ANSWER ALL PARTS
In: Statistics and Probability
8. A pet food company has a business objective of expanding its product line beyond its current kidney- and shrimp-based cat foots. The company developed two new products, one based on chicken liver and the other based on salmon. The company conducted an experiment to compare the two new products with its two existing ones, as well as a generic beef-based product sold in supermarket chains.
For the experiment, a sample of 50 cats from the population at a local animal shelter was selected. Ten cats were randomly assigned to each of the five products being tested. Each of the cats was then presented with 3 ounces of the selected food in a dish at feeding time. The researchers defined the variable to be measured as the number of ounces of food that the cat consumed within a 5-minute period that began when the filled dish was presented to the cat. The results for this experiment are summarized in CatFood.
a. At the 0.05 level of significance, is there evidence of a differences in the mean amount of food eaten among the various products?
b. Does the result in (a) give you statistical permission to probe for individual differences between the food products?
| Kidney | Shrimp | Chicken Liver | Salmon | Beef |
| 3.00 | 0.25 | 2.61 | 0.63 | 2.97 |
| 3.00 | 0.34 | 3.00 | 1.37 | 2.85 |
| 1.47 | 0.16 | 3.00 | 3.00 | 0.44 |
| 2.62 | 0.72 | 2.43 | 0.54 | 0.90 |
| 1.51 | 3.00 | 3.00 | 3.00 | 3.00 |
| 3.00 | 3.00 | 3.00 | 2.55 | 3.00 |
| 3.00 | 0.29 | 2.85 | 2.93 | 0.48 |
| 3.00 | 3.00 | 0.18 | 2.73 | 0.45 |
| 0.74 | 1.99 | 0.98 | 3.00 | 3.00 |
| 2.21 | 2.30 | 1.51 | 0.96 | 0.05 |
In: Statistics and Probability
A prototype automotive tire has a design life of 38,500 miles with a standard deviation of 2,500 miles. The manufacturer tests 60 such tires. On the assumption that the actual population mean is 38,500 miles and the actual population standard deviation is 2,500 miles, find the probability that the sample mean will be less than 36,000 miles. Assume that the distribution of lifetimes of such tires is normal.
(a) Let X = number of miles on a single tire. Write the question above in terms of this variable X.
(b) Using the software tool above, find the probability stated on part (a)
(c) Using the software tool above, graph the probability of stated on part (b)
2. An automobile battery manufacturer claims that its midgrade battery has a mean life of 50 months with a standard deviation of 6 months. Suppose the distribution of battery lives of this particular brand is approximately normal. On the assumption that the claims are true, find the probability that a randomly selected battery of this type will last less than 48 months. (Use the software link for every question)
(a) Let X = number of months a battery will last. Write the question above in terms of this variable X
(b) Find the probability that a single battery of this type will last less than 48 months.
(c) Find the probability that the mean of a random sample of 36 batteries will be less than 48 months.
(d) Why do you think the values from part (b) and part (c) are different? Explain.
In: Statistics and Probability
Exercise 6-50
The accounting department at Weston Materials Inc., a national manufacturer of unattached garages, reports that it takes two construction workers a mean of 30 hours and a standard deviation of 5 hours to erect the Red Barn model. Assume the assembly times follow the normal distribution. Refer to the table in Appendix B.1.
a-1. Determine the z-values for 26 and 35 hours. (Negative answers should be indicated by a minus sign. Round the final answers to 1 decimal place.)
26 hours corresponds to z =
35 hours corresponds to z =
a-2. What percentage of the garages take between 30 hours and 35 hours to erect? (Round the final answer to 2 decimal places.)
Percentage %
b. What percentage of the garages take between 26 hours and 35 hours to erect? (Round the final answer to 2 decimal places.)
Percentage %
c. What percentage of the garages take 25.4 hours or less to erect? (Round the final answer to 2 decimal places.)
Percentage %
d. Of the garages, 10% take how many hours or more to erect? (Round the final answer to 1 decimal place.)
Hours
In: Statistics and Probability
The amount of money spent by Superstore customers is normally distributed with mean 150 and a standard deviation of 12. Suppose that a sample of 64 customers are selected. Answer the following questions.
What is the mean of the sampling distribution?
150
140
170
240
What is the standard error of the sampling distribution?
1.23
1.50
1.55
1.20
What is the probability that the average spending by the customers is greater than 154?
0.9962
0.9966
0.0038
0.0048
What is the probability that the average spending by the customers is less than 154?
0.9962
0.9966
0.0038
0.0048
If the distribution of data is skewed to the left:
Median < Median < Mode
Mean, mode and median are equal
Only mean and mode are equal
None of the above
In: Statistics and Probability