Assuming the sample represents the population very well, that is, the population has approximately the same mean and same standard deviation.
a) 68% of the population fall between ___ inches and ___ inches
b) 95% of the population fall between ___ inches and ___ inches
c) 99.7% of the population fall between ___ inches and ___ inches
Height of 5 yr old Females | |
44.5 | |
45.4 | |
39.6 | |
45.5 | |
42 | |
44.5 | |
39.5 | |
42.3 | |
44 | |
37.7 | |
42.4 | |
43 | |
44.7 | |
43.3 | |
42.2 | |
37.6 | |
43.6 | |
44.6 | |
35.9 | |
34.6 | |
42.2 | |
43.4 | |
36.5 | |
41.2 | |
38.5 | |
42.2 | |
41.3 | |
40.5 | |
43.7 | |
42.4 | |
46.2 | |
44.7 | |
42.7 | |
40.5 | |
43.6 | |
40.3 | |
38.8 | |
37.4 | |
48.1 | |
42.7 | |
45.7 | |
38.6 | |
40.6 | |
44.4 | |
40.6 | |
48.5 | |
41.9 | |
44.5 | |
38.7 | |
47 | |
44.8 | |
41.6 | |
47.5 | |
42.6 | |
45 | |
41.6 | |
40.3 | |
40.7 | |
46 | |
42.8 | |
43.3 | |
50.2 | |
48.4 | |
42 | |
40.7 | |
41.7 | |
42.1 | |
38.2 | |
43.4 | |
39.9 | |
39.5 | |
46.9 | |
37.5 | |
40.3 | |
36.3 | |
38.9 | |
41.9 | |
42.6 | |
44.6 | |
42.3 |
In: Math
Please answer (F)
Let x = age in years of a rural Quebec woman at the time of her first marriage. In the year 1941, the population variance of x was approximately σ2 = 5.1. Suppose a recent study of age at first marriage for a random sample of 31 women in rural Quebec gave a sample variance s2 = 2.7. Use a 5% level of significance to test the claim that the current variance is less than 5.1. Find a 90% confidence interval for the population variance.
(a) What is the level of significance?
State the null and alternate hypotheses.
Ho: σ2 = 5.1; H1: σ2 ≠ 5.1Ho: σ2 = 5.1; H1: σ2 < 5.1 Ho: σ2 < 5.1; H1: σ2 = 5.1Ho: σ2 = 5.1; H1: σ2 > 5.1
(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 binomial population distribution.We assume a normal population distribution. We assume a uniform population distribution.We assume a exponential 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 age at first marriage is less than 5.1.
At the 5% level of significance, there is sufficient evidence to conclude that the that the variance of age at first marriage is less than 5.1.
(f) Find the requested confidence interval for the population
variance. (Round your answers to two decimal places.)
lower limit | |
upper limit |
Interpret the results in the context of the application.
We are 90% confident that σ2 lies above this interval.
We are 90% confident that σ2 lies within this interval.
We are 90% confident that σ2 lies below this interval.
We are 90% confident that σ2 lies outside this interval.
In: Math
The reading speed of second grade students in a large city is approximately normal, with a mean of 90 words per minute (wpm) and a standard deviation of 10 wpm. Complete parts (a) through (f).
(a) What is the probability a randomly selected student in the city will read more than 95 words per minute?
(b) What is the probability that a random sample of 13 second grade students from the city results in a mean reading rate of more than 95 words per minute? then interpret this probability.
(c) What is the probability that a random sample of 26 second grade students from the city results in a mean reading rate of more than 95 words per minute? then interpret.
(d) What effect does increasing the sample size have on the probability? Provide an explanation for this result.
(e) A teacher instituted a new reading program at school. After 10 weeks in the program, it was found that the mean reading speed of a random sample of 20 second grade students was 92.3 wpm. What might you conclude based on this result? Select the correct choice below and fill in the answer boxes within your choice.
(f) There is a 5% chance that the mean reading speed of a random sample of 24 second grade students will exceed what value?
In: Math
Yellow stains on white clothing are stubborn discolorations believed to result from oxidated body oils that have reacted chemically with clothing’s cellulose fibers (as if it were a dye). In as little as one week, there reaction can permanently discolor with clothing with an ugly yellow stain and at this point the most potent laundry product will usually be of little value in removing them. Suppose researchers at the Department of Textiles and Apparel at Cornell University, experimenting with new experimental formulations to remove stains, washed six discolored strips in each of three experimental formulations, and the results as follows Coloring rating (after washing) 0 = Dark Yellow, the initial shade 20 = Pure White-All washed strips were matched to a color chart, a series of increasingly lighter shade, and rated from 0 to 20. Formulation X Y Z 10 8 11 9 10 12 10 12 14 12 9 11 8 10 10 11 8 14 Three Different Formulations were tested a) Calculate by hand the average color rating and variance for each group b) Assuming valid random samples, test at 5% level significance whether differences among the three-sample means can be attributed to chance fluctuation or, indeed, to actual differences in the stain-eliminating ability of the three formulations. ( hint: find the F-ratio) c) Briefly discuss the implications of these results. Need to solve by hand. Please help!
In: Math
A research council wants to estimate the mean length of time (in minutes) that the average U.S. adult spends watching television using digital video recorders (DVR’s) each day. To determine the estimate, the research council takes random samples of 35 U.S. adults and obtains the following times in minutes.
24 |
27 |
26 |
29 |
33 |
21 |
18 |
24 |
23 |
34 |
17 |
15 |
19 |
23 |
25 |
29 |
36 |
19 |
18 |
22 |
16 |
45 |
32 |
12 |
24 |
35 |
14 |
40 |
30 |
19 |
14 |
28 |
32 |
15 |
39 |
From past studies, the research council has found that the standard deviation time is 4.3 minutes and that the population of times is normally distributed.
Construct a 90% confidence interval for the population mean.
Construct a 99% confidence interval for the population mean.
Interpret the results and compare the widths of the confidence intervals.
Test the claim that the mean time spent watching DVR’s is 20 minutes each day using a significance level of 0.05.
You may use Stat Disk, TI-84 calculator, or CrunchIt to find the confidence intervals. Be sure to show your results in your post.
In: Math
(3) A random sample of 51 fatal crashes in 2009 in which the driver had a positive blood alcohol concentration (BAC) from the National Highway Traffic Safety Administration results in a mean BAC of 0.167 grams per deciliter (g/dL) with a standard deviation of 0.010 g/dL.
(a) What is a point estimate for the mean BAC of all fatal crashes with a positive BAC?
(b) In 2009, there were approximately 25,000 fatal crashes in which the driver had a positive BAC. Explain why this, along with the fact that the data were obtained using a simple random sample, satisfies the requirements for constructing a confidence interval.
(c) Determine and interpret a 98% confidence interval for the mean BAC in fatal crashes in which the driver had a positive BAC. Do not use the T-Interval or Z-Interval features of your calculator.
In: Math
8. Given the following data: Number of patients = 3,293 Number of patients who had a positive test result and had the disease = 2,184 Number of patients who had a negative test, and did not have the disease = 997 Number of patients who had a positive test result, but did not have the disease = 55 Number of patients who had a negative test result, but who had the disease = 57 a. Create a complete and fully labeled 2x2 table (10) b. Calculate the positive predictive value (5) c. Calculate the negative predictive value (5) d. Calculate the likelihood ratio for a positive test result (5) e. Calculate the likelihood ratio for a negative test result (5)
Show all work
In: Math
Jan Northcutt, owner of Northcutt Bikes, started business in 1995. She notices the quality of bikes she purchased for sale in her bike shop declining while the prices went up. She also found it more difficult to obtain the features she wanted on ordered bikes without waiting for months. Her frustration turned to a determination to build her own bikes to her particular customer specifications.
She began by buying all the necessary parts (frames, seats, tires, etc.) and assembling them in a rented garage using two helpers. As the word spread about her shop’s responsiveness to options, delivery, and quality, however, the individual customer base grew to include other bike shops in the area. As her business grew and demanded more of her attention, she soon found it necessary to sell the bike shop itself and concentrate on the production of bikes from a fairly large leased factory space.
As the business continued to grow, she backward integrated more and more processes into her operation, so that now she purchases less than 50% of the component value of the manufactured bikes. This not only improves her control of production quality but also helps her control the costs of production and makes the final product more cost attractive to her customers.
The Current Situation
Jan considers herself a hands-on manager and has typically used her intuition and her knowledge of the market to anticipate production needs. Since one of her founding principles was rapid and reliable delivery to customer specification, she felt she needed to begin production of the basic parts for each particular style of bike well in advance of demand. In that way she could have the basic frame, wheels, and standard accessories started in production prior to the recognition of actual demand, leaving only the optional add-ons to assemble once the order came in. Her turnaround time for an order of less than half the industry average is considered a major strategic advantage, and she feels it is vital for her to maintain or even improve on response time if she is to maintain her successful operation.
As the customer base have grown, however, the number of customers Jan knows personally has shrunk significantly as a percentage of the total customer base for Northcutt Bikes, and many of these new customers are expecting or even demanding very short response times, as that is what attracted them to Northcutt Bikes in the first place. This condition, in addition to the volatility of overall demand, has put a strain on capacity planning. She finds that at times there is a lot of idle time (adding significantly to costs), whereas at other times the demand exceeds capacity and hurts customer response time. The production facility has therefore turned to trying to project demand for certain models, and actually building a finished goods inventory of those models. This has not proven to be too satisfactory, as it has actually hurt costs and some response times. Reasons include the following:
- The finished goods inventory is often not the “right” inventory, meaning shortages for some goods and excessive inventory of others. This condition both hurts responsiveness and increases inventory costs.
- Often, to help maintain responsiveness, inventory is withdrawn from finished goods and reworked, adding to product cost.
- Reworking inventory uses valuable capacity for other customer orders, again resulting in poorer response times and/or increased costs due to expediting. Existing production orders and rework orders are both competing for vital equipment and resources during times of high demand, and scheduling has become a nightmare.
The inventory problem has grown to the point that additional storage space is needed, and that is a cost that Jan would like to avoid if possible.
Another problem that Jan faces is the volatility of demand for bikes. Since she is worried about unproductive idle time and yet does not wish to lay off her workers during times of low demand, she has allowed them to continue to work steadily and build finished goods. This makes the problem of building the “right” finished goods even more important, especially given the tight availability of storage space.
Past Demand
The following shows the monthly demand for one major product line: the standard 26-inch 10-speed street bike. Although it is only one of Jan’s products, it is representative of most of the major product lines currently being produced by Northcutt Bikes. If Jan can find a way to sue this data to more constructively understand her demand, she feels she can probably use the same methodologies to project demand for other major product families. Such knowledge can allow her, she feels, to plan more effectively and continue to be responsive while still controlling costs.
Actual Demand |
||||
Month |
2011 |
2012 |
2013 |
2014 |
January |
437 |
712 |
613 |
701 |
February |
605 |
732 |
984 |
1291 |
March |
722 |
829 |
812 |
1162 |
April |
893 |
992 |
1218 |
1088 |
May |
901 |
1148 |
1187 |
1497 |
June |
1311 |
1552 |
1430 |
1781 |
July |
1055 |
927 |
1392 |
1843 |
August |
975 |
1284 |
1481 |
839 |
September |
822 |
1118 |
940 |
1273 |
October |
893 |
737 |
994 |
912 |
November |
599 |
983 |
807 |
996 |
December |
608 |
872 |
527 |
792 |
1. Plot the data and describe what you see. What does it mean and how would you use the information from the plot to help you develop a forecast?
2. Use at least two different methodologies to develop as accurate a forecast as possible for the demand. Use each of those methods to project the next four months demand.
3. Which method from question 2 is “better”? How do you know that?
In: Math
Nitterhouse Masonry Products, LLC, in Chambersburg, Pennsylvania, produces architectural concrete masonry products. The Dover, the largest block in a certain collection, is used primarily for residential retaining walls, and is manufactured to weigh 45 pounds. A quality control inspector for the company randomly selected 17 blocks, and determined that they have an average weight of 46.6 pounds with a sample standard deviation of 3.20 pounds. Assume that the distribution of the weights of the blocks is normal. Please use 4 decimal places for all critical values.
(0.5 pts.) a) Should a z or t distribution be used for statistical procedures regarding the mean? Please explain your answer.
b) Is there any evidence to suggest that the true mean weight is
not 45 pounds at a 5% significance level?
Calculate the test statistic
Calculate the p-value.
Write the complete four steps of the hypothesis test below. The work for all parts will be at the end of the question.
c) Calculate the 95% confidence interval for the mean.
d) Explain why parts b) and c) state the same thing. That is, what in part b) is consistent with what in part c)?
e) Is there strong evidence for your decision of "reject the null hypothesis" or "fail to reject the null hypothesis"? Please explain your answer using the results from both the hypothesis test and the confidence interval.
In: Math
he number of chocolate chips in an 18-ounce bag of chocolate chip cookies is approximately normally distributed with a mean of 1252 chips and standard deviation 129 chips.
(a) What is the probability that a randomly selected bag contains between 1100 and 1400 chocolate chips, inclusive?
(b) What is the probability that a randomly selected bag contains fewer than 1025 chocolate chips?
(c) What proportion of bags contains more than 1200 chocolate chips?
(d) What is the percentile rank of a bag that contains 1050 chocolate chips?
In: Math
Commercial real estate prices and rental rates suffered substantial declines in 2008 and 2009.† These declines were particularly severe in Asia; annual lease rates in Tokyo, Hong Kong, and Singapore declined by 40% or more. Even with such large declines, annual lease rates in Asia were still higher than those in many cities in Europe. Annual lease rates for a sample of 30 commercial properties in an Asian city showed a mean of $1,118 per square meter with a standard deviation of $230. Annual lease rates for a sample of 40 commercial properties in a European city showed a mean lease rate of $987per square meter with a standard deviation of $195.
b) what is the value of the test statistic? (round your answer to three decimal places)
c) what is the p-value? (round your answer to four decimal places.)
In: Math
Chi Square Test
We will now use Excel to run an example of a chi square test.
Chi square test is checking the independence of two variables. Our
example will test if taking hormonal pills and being overweight are
related. We will test the independence on 200 random patients.
Thus, N=200. They will be divided first into two groups, those who
take hormonal pills and those who do not. Second, they will be
divided into three groups based on weight, not overweight,
overweight and obese. All data is in this table
Observed frequency table Not overweight
overweight obese total
Not taking hormonal pills 35 36 49
120
Take hormonal pills 33 32 15 80
Total 68 68 64 200
We will start in Excel by making the above table in region
A1-E4, first five columns and first four rows. That is, in celll B1
you will type Not overweight, in cell A2 Not taking hormonal pills,
etc
Next we construct the expected table. Let's make it in the region
A8-E11. Type Expected frequency table in cell A8, not overweight in
cell B8 etc. Data in the table is calculated in this fashion. Cell
B10 corresponds to the take hormonal pills row and not overweight
column. Thus in cell B10 we type =B4*E3/E4. In cell D10 we type
=D4*E3/E4. Using that strategy complete the expected frequency
table.
Next we check if chi square test will work for this example. When
you remove total from the expected frequency table, you have a 2x3
table with 6 entries. To run chi square we should first have no
zero entries out of those 6. In cell A13 type zero entries. In cell
B13 type the actual value of how many zero entries you have in
expected frequency table. Second, you should have at most 20%
entries that are less than 5. In cell A14 type percentage of
entries less than 5. In cell B14 calculate the actual value of
percents of entries in expected frequency table that are less than
5.
Now let's evaluate chi square parameters. In cell A16 type df. In
cell B16 evaluate df. In cell A20 type chi square. We will evaluate
chi square in cell B20. In cell B20 type
=(B2-B9)^2/B9+(B3-B10)^2/B10+(C2-C9)^2/C9+(C3-C10)^2/C10+(D2-D9)^2/D9+(D3-D10)^2/D10.
In cell A22 type table chi square and then find the table value on
page 416 with .05 level of significance and degrees of freedom df
from B16. Put that value in cell B22.
Now we do testing. In cell A24 type H0 and in cell B24 state the
null hypothesis. In cell A25 type H1 and in cell B25 state the
alternate hypothesis.
Now compare the values in cells B20 and B22. State if we reject or
do not reject the null hypothesis in cell A26. Explain how you
obtained your conclusion in cell B26.
Next we will test it another way, with asymptotic significance
(probability).
In cell A28 type Asymp. Sig. (probability). We will evaluate Sig.
in cell B28. We will use an Excel command for finding sig. in a chi
square test. In cell B28 type =CHITEST(B2:D3,B9:D10).
Compare the sig. in cell B28 with the significance level of .05 and
using that comparison, state in cell A31 if we reject or do not
reject the null hypothesis. Explain how you have reached your
statement in cell B31.
In: Math
Is 2k-1 odd?
I get that 2(some int k) + 1 is the property for odd numbers.
The main question:
I am confused on how 2k-1= 2k-2+1 which is a form of k?
In: Math
Height | Weight | Age | Shoe Size | Waist Size | Pocket Change |
64 | 180 | 39 | 7 | 36 | 18 |
66 | 140 | 31 | 9 | 30 | 125 |
69 | 130 | 31 | 9 | 25 | 151 |
63 | 125 | 36 | 7 | 25 | 11 |
68 | 155 | 24 | 8 | 31 | 151 |
62 | 129 | 42 | 6 | 32 | 214 |
63 | 173 | 30 | 8 | 34 | 138 |
60 | 102 | 26 | 6 | 25 | 67 |
66 | 180 | 33 | 8 | 30 | 285 |
66 | 130 | 31 | 9 | 30 | 50 |
63 | 125 | 32 | 8 | 26 | 32 |
68 | 145 | 33 | 10 | 28 | 118 |
75 | 235 | 44 | 12 | 40 | 60 |
68 | 138 | 43 | 8 | 27 | 50 |
65 | 165 | 55 | 9 | 30 | 22 |
64 | 140 | 24 | 7 | 31 | 95 |
78 | 240 | 40 | 9 | 38 | 109 |
71 | 163 | 28 | 7 | 32 | 14 |
68 | 195 | 24 | 10 | 36 | 5 |
66 | 122 | 33 | 9 | 26 | 170 |
53 | 115 | 25 | 7 | 25 | 36 |
71 | 210 | 30 | 10 | 36 | 50 |
78 | 108 | 23 | 7 | 22 | 75 |
69 | 126 | 23 | 8 | 24 | 175 |
77 | 215 | 24 | 12 | 36 | 41 |
68 | 125 | 23 | 8 | 30 | 36 |
62 | 105 | 50 | 6 | 24 | 235 |
69 | 126 | 42 | 9 | 27 | 130 |
55 | 140 | 42 | 8 | 29 | 14 |
67 | 145 | 30 | 8 | 30 | 50 |
1. weight vs. age α ̇=.01/2
Step 1: Ho: __ _ ___
Ha: __ _ ___
Step 2:
Alpha level = _____
Step 3:
Sampling
distribution is df = _____
Step 4:
Decision
Rule: I will reject the Ho if the |_robs_| value falls at or
beyond
the |_rcrit_| of ____, otherwise I will fail to reject
Step 5:
Calculation:
\_robs_/ = _____
Step 6: Summary: Since the |_robs_| of ____ _____________ the |_rcrit_| of
_____, I therefore reject/fail to reject (choose one) the
Ho.
Step 7: Conclusion: Since _______ occurred, I conclude ___________________________________________________________________.
2. height vs. shoe size α ̇=.02/2
Step
1:
Ho: __ _ ___
Ha: __ _ ___
Step 2:
Alpha level = _____
Step 3:
Sampling
distribution is df = _____
Step 4:
Decision
Rule: I will reject the Ho if the |_robs_| value falls at or
beyond
the |_rcrit_| of ____, otherwise I will fail to reject
Step 5:
Calculation:
\_robs_/ = _____
Step 6: Summary: Since the |_robs_| of ____ _____________ the |_rcrit_| of
_____, I
therefore reject/fail to reject (choose one) the Ho.
Step 7: Conclusion: Since _______ occurred, I conclude ___________________________________________________________________.
3.Explain the correlation coefficient of determination.
In: Math
Match each example to the type of bias that would result
|
|
In: Math