The professors who teach the Introduction to Psychology course at State University pride themselves on the normal distributions of exam scores. After the first exam, the current professor reports to the class that the mean for the exam was 73, with a standard deviation of 7.
a. What proportion of student would be expected to score above 80?
b What proportion of students would be expected to score between 55 and 75?
c. What proportion of students would be expected to score less than 65?
d. If the top 10% of the class receive an A for the exam, what score would be required for a student to receive an A?
e. If the bottom 10% of the class fail the exam, what score would earn a student a failing grade?
In: Math
using chapter 13 data set 2, the researchers want to find out whether there is a difference among the graduation rates (and these are percentages) of five high schools over a 10-year period. Is there? (hint: are the years a factor?)
High School 1 | High School 2 | High School 3 | High School 4 | High School 5 | |
2003 | 67 | 82 | 94 | 65 | 88 |
2004 | 68 | 87 | 78 | 65 | 87 |
2005 | 65 | 83 | 81 | 45 | 86 |
2006 | 68 | 73 | 76 | 57 | 88 |
2007 | 67 | 77 | 75 | 68 | 89 |
2008 | 71 | 74 | 81 | 76 | 87 |
2009 | 78 | 76 | 79 | 77 | 81 |
2010 | 76 | 78 | 89 | 72 | 78 |
2011 | 72 | 76 | 76 | 69 | 89 |
2012 | 77 | 86 | 77 | 58 | 87 |
In: Math
To study the effect the ecological impact of malaria, researchers in California measured the effect of malaria on what distance an animal could run in 2 minutes. They used a sample of a local lizard species, Sceloporis occidentalis, collected in the field. They selected 15 lizards from those found to be infected with the malarial parasite Plasmodium mexicanum and 15 lizards found not infected.
The distance each lizard could run in a time limit of 2 minutes was recorded in a controlled environment. The data (below) is also in an Excel file, “Malaria effect”
a) Does this design use paired data or independent samples? Explain.
b) Compute a 95% confidence interval for the difference in distance ran between the two malarial infections of the lizards. Show all of your working.
c) Based on this confidence interval, can you conclude there is a difference in mean distance ran between the infected and uninfected lizards?
Infected |
16.4 |
29.4 |
37.1 |
23.0 |
24.1 |
24.5 |
16.4 |
29.1 |
36.7 |
28.7 |
30.2 |
21.8 |
37.1 |
20.3 |
28.3 |
Uninfected |
22.2 |
34.8 |
42.1 |
32.9 |
26.4 |
30.6 |
32.9 |
37.5 |
18.4 |
27.5 |
45.5 |
34.0 |
45.5 |
24.5 |
28.7 |
d) Conduct a test of significance (at a significance level of 5%) on the difference between infected and uninfected lizards to decide whether the uninfected lizards can run a further distance on average. Follow all steps clearly and write a clear conclusion.
e) Do you have the same conclusion from the confidence interval calculation in c) as that concluded by the test done in d) above? Explain why or why not.
In: Math
The U.S. Energy Information Administration (US EIA) reported that the average price for a gallon of regular gasoline is $2.94. The US EIA updates its estimates of average gas prices on a weekly basis. Assume the standard deviation is $0.20 for the price of a gallon of regular gasoline and recommend the appropriate sample size for the US EIA to use if they wish to report each of the following margins of error at 95% confidence. (Round your answers up to the nearest whole number.)
(a) The desired margin of error is $0.10. ______
(b) The desired margin of error is $0.06. _________
(c) The desired margin of error is $0.05. ________
In: Math
A study was
done on body temperatures of men and women. The results are shown
in the table. Assume that the two samples are independent simple
random samples selected from normally distributed populations, and
do not assume that the population standard deviations are equal.
Complete parts (a) and (b) below. Use a
0.010.01 significance level for both parts. |
Men |
Women |
|||
muμ |
mu 1μ1 |
mu 2μ2 |
|||
n |
1111 |
5959 |
|||
x overbarx |
97.6997.69degrees°F |
97.2597.25degrees°F |
|||
s |
0.920.92degrees°F |
0.710.71degrees°F |
FIND P VALUE
T VALUE
CONFIDENCE INTERVAL FOR TESTING MEN HAVE A HIGER BODY TEMP THAN WOMEN
In: Math
A 95% confidence interval for a population mean was reported to be 148.79 to 155.21. If σ = 15, what sample size was used in this study? (Round your answer to the nearest integer.)
In: Math
A support used in an automotive application is supposed to have a nominal internal diameter of 1.5 inches. A random sample of 25 brackets is selected and the nominal internal diameter of these brackets is 1.4975 inches. The diameters of the supports are known to be normally distributed with a standard deviation of σ = 0.01 inches.
a) Perform the hypothesis test Ho: μ = 1.5 versus H1 ≠ 1.5 at an α = 0.01 using the 5 hypothesis test steps: 1. Establish the hypotheses
Ho:
H1:
2. Establish the test statistic
3. Establish the rejection zone Rejection Zone:
Graph showing rejection zone (Place the value on the vertical line where the rejection zone begins):
4. Calculate test statistics
5. Establish the conclusion Graph showing rejection zone and placement of test statistics. (Place the value on the vertical line that corresponds to the test statistic):
Based on the previous graph, establish the conclusion that applies to the problem.
b) Determine the p-value (probability of the test statistic) The p-value depends on whether the test is one-tailed or two-tailed. If the hypothesis test is of a tail, then the p-value is the value that comes directly from the probability of the test statistic. If the test is two-tailed, then the p-value is 2 times the probability obtained from looking for the probability of the test statistic
. The value of Ф (1.25) =
The value of α / 2 =
P-value =
c) Based on the p-value and considering an α = 0.01, what would be the decision? Reject or not reject Ho? Use the space to answer.
d) Use the following OC graph to approximate the probability of Type II Error for a true average diameter of 1,495. Start by calculating the value of parameter d with the following formula.
? = ⌈μ − μ0⌉? =
The value of Type II Error is approximately: β = approximately
The "Power" of the test is: Power = 1 - β = approximately
e) What sample size is required to detect a true average diameter as low as 1,495 inches if we want the “Power” of the test to be 90%? Start by determining how much Error Type II of the “Power” = 1 - β ratio is.
β =
? =
⌈μ − μ0⌉? = n ≈
In: Math
Central Tendency & Descriptive Statistics:
You have colleagues who reside in different zip codes. Which measure(s) of central tendency (mean, median, mode) or other descriptive statistics would you use to describe this information? (please answer in full paragraph - thank you)
In: Math
The mean high temperature on the equatorial island of Pacifica is µ = 82 F. For a random sample of 30 days following the eruption of a volcano on another island the mean high temperature on Pacifica was 76 F with standard deviation 8 . Does the data indicate that the mean temperature on Pacifica has changed? Use a 5% significance level.
In: Math
Suppose a sample of 1042 tenth graders is drawn. Of the students sampled, 865 read above the eighth grade level. Using the data, construct the 90% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level. Round your answers to three decimal places.
I need help figuring out the formula for this problem.
In: Math
sp shift success
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Belinda Daytime N
Belinda Daytime N
Belinda Daytime N
Belinda Daytime N
Belinda Daytime N
Belinda Daytime N
Belinda Daytime N
Belinda Daytime N
Belinda Daytime N
Belinda Daytime N
Belinda Daytime N
Belinda Daytime N
Belinda Daytime N
Belinda Daytime N
Belinda Daytime N
Belinda Daytime N
Belinda Daytime N
Belinda Daytime N
Belinda Daytime N
Belinda Daytime N
Belinda Daytime N
Belinda Daytime N
Belinda Daytime N
Belinda Daytime N
Belinda Daytime N
Belinda Daytime N
Belinda Daytime N
Belinda Daytime N
Belinda Daytime N
Belinda Daytime N
Belinda Daytime N
Belinda Daytime N
Belinda Daytime N
Belinda Daytime N
Belinda Daytime N
Belinda Daytime N
Belinda Daytime N
Belinda Daytime N
Belinda Daytime N
Belinda Daytime N
Belinda Daytime N
Belinda Daytime N
Belinda Daytime N
Belinda Daytime N
Belinda Daytime N
Belinda Daytime N
Belinda Daytime N
Belinda Daytime N
Belinda Daytime N
Belinda Daytime N
Belinda Daytime N
Belinda Daytime N
Belinda Daytime N
Belinda Daytime N
Belinda Daytime N
Belinda Daytime N
Belinda Daytime N
Belinda Daytime N
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Belinda Daytime N
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Belinda Daytime N
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Belinda Daytime N
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Belinda Daytime N
Belinda Daytime N
Belinda Daytime N
Belinda Daytime N
Belinda Daytime N
Belinda Daytime N
Belinda Daytime N
Belinda Daytime N
Belinda Daytime N
Belinda Daytime N
Belinda Daytime N
Belinda Daytime N
Belinda Daytime N
Belinda Daytime N
Belinda Daytime N
Belinda Daytime N
Belinda Daytime N
Belinda Daytime N
Belinda Daytime N
Belinda Daytime N
Belinda Daytime N
Belinda Daytime N
Belinda Daytime N
Belinda Daytime N
Belinda Daytime N
Belinda Daytime N
Belinda Daytime N
Belinda Daytime N
Belinda Daytime N
Belinda Daytime N
Belinda Daytime N
Belinda Daytime N
Belinda Daytime N
Belinda Daytime N
Belinda Daytime N
Belinda Daytime N
Belinda Daytime N
Belinda Daytime N
Belinda Daytime N
Belinda Daytime N
Belinda Daytime N
Belinda Evening N
Belinda Evening N
Belinda Evening N
Belinda Evening N
Belinda Evening N
Belinda Evening N
Belinda Evening N
Belinda Evening N
Belinda Evening N
Belinda Evening N
Belinda Evening N
Belinda Evening N
Belinda Evening N
Construct a contingency table to summarize the distribution of
transactions by sales-
person (Alphonse, Belinda) and by success (Y, N). Please include
the margins in your
table.
(a) (3 points) First express the contingency table as a frequency
table (in terms of
counts)
(b) (1 point) Reproduce the table only this time in terms of
relative frequency (i.e.,
proportions vis a vis the total, rather than row-totals or
column-totals)
(c) Zihan, PJs chief data analyst, has randomly chosen one
transaction from the dataset
represented by the contingency table above (i.e., part b). Please
calculate the
following.
i. (2 points) The probability that the transaction is attributed to
Alphonse (re-
gardless of success) or represents a successful sale attributed to
Belinda.
ii. (2 points) Zihan informs you that the transaction represents a
successful sale.
Calculate the probability that the sale was made by Belinda.
In: Math
Quality Specifications for the bottle filling process = 355 ± 1.5 ml
The sample measurements for Process A & Process B can be found in the attached Excel file Other relevant information for the analysis:
Process A = 11.5 million bottles
Process B = 6.9 million bottles
Estimated cost of overfilling = $0.071 per bottle
Estimated cost of underfilling = $0.134 per bottle
Process A | Process B | ||||
bottle nbr | fill volume, ml | bottle nbr | fill volume, ml | ||
1 | 353.8716 | 1 | 356.4036 | ||
2 | 356.4629 | 2 | 354.8854 | ||
3 | 354.3566 | 3 | 356.2884 | ||
4 | 354.9326 | 4 | 355.9886 | ||
5 | 354.1558 | 5 | 355.3441 | ||
6 | 354.6894 | 6 | 355.141 | ||
7 | 353.1613 | 7 | 355.5605 | ||
8 | 354.492 | 8 | 354.7924 | ||
9 | 353.2064 | 9 | 355.7594 | ||
10 | 355.0353 | 10 | 356.2499 | ||
11 | 354.1497 | 11 | 356.7416 | ||
12 | 355.3837 | 12 | 355.6718 | ||
13 | 354.8073 | 13 | 355.8648 | ||
14 | 354.52 | 14 | 354.8881 | ||
15 | 354.127 | 15 | 355.7184 | ||
16 | 354.0073 | 16 | 355.5325 | ||
17 | 354.5865 | 17 | 355.5129 | ||
18 | 355.2267 | 18 | 356.0567 | ||
19 | 354.1048 | 19 | 355.9526 | ||
20 | 354.7092 | 20 | 356.2481 | ||
21 | 354.8251 | 21 | 355.5009 | ||
22 | 355.1323 | 22 | 355.8066 | ||
23 | 355.6323 | 23 | 355.6735 | ||
24 | 355.0618 | 24 | 355.5141 | ||
25 | 355.6289 | 25 | 355.6569 | ||
26 | 354.5315 | 26 | 355.0254 | ||
27 | 354.6454 | 27 | 356.4625 | ||
28 | 354.1473 | 28 | 356.3046 | ||
29 | 355.5054 | 29 | 356.2273 | ||
30 | 354.6658 | 30 | 355.4353 | ||
31 | 354.777 | 31 | 356.0174 | ||
32 | 354.6489 | 32 | 356.3742 | ||
33 | 354.9304 | 33 | 355.3607 | ||
34 | 356.0081 | 34 | 355.6758 | ||
35 | 353.5191 | 35 | 355.2479 | ||
36 | 354.089 | 36 | 356.5349 | ||
37 | 355.4005 | 37 | 356.1038 | ||
38 | 354.7968 | 38 | 356.1314 | ||
39 | 354.5803 | 39 | 356.1499 | ||
40 | 354.5402 | 40 | 356.9204 | ||
41 | 353.9612 | 41 | 356.0494 | ||
42 | 355.3751 | 42 | 355.8082 | ||
43 | 355.2035 | 43 | 355.7958 | ||
44 | 354.7033 | 44 | 356.4498 | ||
45 | 355.5842 | 45 | 355.0805 | ||
46 | 355.2069 | 46 | 355.6821 | ||
47 | 355.291 | 47 | 355.7853 | ||
48 | 355.5132 | 48 | 356.0396 | ||
49 | 354.4062 | 49 | 354.5163 | ||
50 | 354.8773 | 50 | 355.1708 | ||
51 | 354.0812 | 51 | 356.8699 | ||
52 | 355.5711 | 52 | 355.8047 | ||
53 | 356.8612 | 53 | 356.1663 | ||
54 | 354.6389 | 54 | 356.2781 | ||
55 | 355.4831 | 55 | 355.6501 | ||
56 | 354.4165 | 56 | 355.1498 | ||
57 | 354.4106 | 57 | 356.1733 | ||
58 | 353.8034 | 58 | 355.3848 | ||
59 | 355.779 | 59 | 355.4425 | ||
60 | 354.3574 | 60 | 355.7853 | ||
61 | 354.2061 | 61 | 355.6234 | ||
62 | 355.3 | 62 | 355.2701 | ||
63 | 353.8064 | 63 | 355.3693 | ||
64 | 355.0172 | 64 | 356.0998 | ||
65 | 355.2049 | 65 | 354.3443 | ||
66 | 356.0506 | 66 | 355.2375 | ||
67 | 355.5254 | 67 | 356.0556 | ||
68 | 355.9298 | 68 | 355.6644 | ||
69 | 354.6942 | 69 | 355.9695 | ||
70 | 354.879 | 70 | 356.0207 | ||
71 | 354.876 | 71 | 355.8412 | ||
72 | 353.2011 | 72 | 356.013 | ||
73 | 355.69 | 73 | 356.0578 | ||
74 | 355.2879 | 74 | 355.0693 | ||
75 | 354.881 | 75 | 356.2371 | ||
76 | 353.4271 | 76 | 356.4531 | ||
77 | 354.3281 | 77 | 355.8708 | ||
78 | 355.4182 | 78 | 355.7516 | ||
79 | 354.7104 | 79 | 356.0311 | ||
80 | 354.5383 | 80 | 355.336 | ||
81 | 354.2397 | 81 | 356.6274 | ||
82 | 355.7615 | 82 | 355.5591 | ||
83 | 355.7941 | 83 | 355.577 | ||
84 | 353.7047 | 84 | 356.3873 | ||
85 | 355.3057 | 85 | 355.4378 | ||
86 | 355.4152 | ||||
87 | 355.7074 | ||||
88 | 354.8495 | ||||
89 | 356.3219 | ||||
90 | 355.2006 | ||||
91 | 356.162 | ||||
92 | 356.7196 | ||||
93 | 354.69 | ||||
94 | 354.7049 | ||||
95 | 355.2266 | ||||
96 | 355.7611 | ||||
97 | 356.1532 | ||||
98 | 355.2149 | ||||
99 | 354.9555 | ||||
100 | 356.2889 | ||||
101 | 356.1144 | ||||
102 | 355.8599 | ||||
103 | 356.2266 | ||||
104 | 356.2091 | ||||
105 | 355.8744 | ||||
106 | 355.8112 | ||||
107 | 355.1513 | ||||
108 | 355.2167 | ||||
109 | 355.2743 | ||||
110 | 355.2112 | ||||
111 | 355.8082 | ||||
112 | 355.8028 | ||||
a. Calculate numerical measures for Central Tendency and for Dispersion on both processes.
b. Construct confidence interval for the true Mean (µ) of each of the processes. Comment on the results. Is there an issue with the mean of any of the processes? Explain the results.
c. Construct confidence interval for the true Standard Deviation (σ) of each of the processes. Comment on the results. Is there an issue with the dispersion of any of the processes? Explain the results.
d. Run a hypothesis test (a t-test) for the Mean of each process being equal to the target value of 355 ml. Comment on the results e. Draw a histogram (with normal-distribution fit) for both processes; interpret the results (also, draw the two theoretical normal distributions overlapping in the same graph to facilitate interpretation)
f. Construct a Normal Probability Plot for each of the process. What can you conclude?
g. Estimate the expected number of bottles overfilled per year from each of the processes
h. Estimate the expected number of bottles overfilled per year from each of the processes
i. Calculate and compare the annual cost of overfilling AND underfilling per process. Comment on the results.
j. Make your Final Conclusions and Recommendations
minitab plz
In: Math
Suppose that the distribution for total amounts spent by students vacationing for a week in Florida is normally distributed with a mean of 650 and a standard deviation of 120. Suppose you take a SRS of 15 students from this distribution. What is the probability that an SRS of 15 students will spend an average of between 600and700? Round to five decimal places.
In: Math
Both data sets have a mean of 245. One has a standard deviation of 16, and the other has a standard deviation of 24. LOADING... Click the icon to view data sets. Which data set has which deviation?
Bold left parenthesis a right parenthesis(a) |
2020 |
88 |
99 |
Key:
2020 |88equals=208208 |
Bold left parenthesis b right parenthesis(b) |
2020 |
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2121 |
44 |
55 |
88 |
2121 |
55 |
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2222 |
11 |
55 |
2222 |
22 |
33 |
55 |
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2323 |
00 |
00 |
66 |
77 |
2323 |
00 |
22 |
55 |
88 |
88 |
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2424 |
44 |
55 |
66 |
2424 |
11 |
11 |
22 |
33 |
33 |
33 |
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2525 |
11 |
33 |
66 |
88 |
2525 |
11 |
55 |
88 |
88 |
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2626 |
00 |
55 |
99 |
2626 |
22 |
33 |
44 |
55 |
|||||||||
2727 |
66 |
2727 |
00 |
55 |
|||||||||||||
2828 |
33 |
55 |
77 |
2828 |
A. (a) has a standard deviation of 24 and (b) has a standard deviation of 16, because the data in (a) have more variability.
B. (a) has a standard deviation of 16 and (b) has a standard deviation of 24, because the data in (b) have less variability.
In: Math
Sleeping outlier: A simple random sample of nine college freshmen were asked how many hours of sleep they typically got per night. The results were
8.5 24 9 8.5 8 6.5 6 7.5 9.5
Notice that one joker said that he sleeps 24 a day.
(a) The data contain an outlier that is clearly a mistake. Eliminate the outlier, then construct a 99.9% confidence interval for the mean amount of sleep from the remaining values. Round the answers to at least two decimal places.
(b) Leave the outlier in and construct the 99.9% confidence interval. Round the answers to at least two decimal places.
(c) Are the results noticeably different?
In: Math