Slices of pizza for a certain brand of pizza have a mass that is approximately normally distributed with a mean of 66.8 grams and a standard deviation of 1.94 grams.
a) For samples of size 16 pizza slices, what is the standard deviation for the sampling distribution of the sample mean?
b) What is the probability of finding a random slice of pizza with a mass of less than 66.3 grams?
c) What is the probability of finding a 16 random slices of pizza with a mean mass of less than 66.3 grams?
d) What sample mean (for a sample of size 16) would represent the bottom 15% (the 15th percentile)? grams
In: Math
A researcher examined whether the time of day someone exercises affects memory retention in college courses. Participants were assigned to one of three exercise groups: morning, afternoon, evening, and their performance on a memorization task was measured. This data is below:
| Morning | Afternoon | Evening |
| 6 | 4 | 7 |
| 7 | 5 | 6 |
| 8 | 6 | 8 |
| 5 | 4 | 6 |
| 6 | 5 | 5 |
| Mean = 6.40 | Mean = 4.80 | Mean = 6.40 |
| s = 1.14 | s =.84 | s = 1.14 |
Assuming the researcher wants to know whether the performance was different in these groups (alpha = .05). What is the Sum of Squares Between (SSB)? Round to two decimal places.
In: Math
2 The Civil War. A national survey conducted among a simple random sample of 1,507 adults shows that 56% of Americans think the Civil War is still relevant to American politics and political life.
(a) Conduct a hypothesis test to determine if these data provide strong evidence that the majority of the Americans think the Civil War is still relevant.
(b) Interpret the p-value in this context.
(c) Calculate a 90% confidence interval for the proportion of Americans who think the Civil War is still relevant. Interpret the interval in this context, and comment on whether or not the confidence interval agrees with the conclusion of the hypothesis test
In: Math
In a sample of 1000 randomly selected consumers who had opportunities to send in a rebate claim form after purchasing a product, 240 of these people said they never did so. Reasons cited for their behavior included too many steps in the process, amount too small, missed deadline, fear of being placed on a mailing list, lost receipt, and doubts about receiving the money.
Calculate an upper confidence bound at the 95% confidence level for the true proportion of such consumers who never apply for a rebate. (Round your answer to four decimal places.)
In: Math
8) Scores on an exam have a normal distribution with a mean of 80 and a standard deviation of 12.
a) Find the probability that a person would score above 90.
b) Find the probability that a person would score between 75 and 85.
c) Find the probability that a group of 7 people would have a mean score above 84.
d) Find the score needed to be in the top 10% of the class.
In: Math
For each combination of sample size and sample proportion, find the approximate margin of error for the 95% confidence level. (Round the answers to three decimal places.) (a) n = 200, p̂ = 0.53. (b) n = 500, p̂ = 0.53. (c) n = 500, p̂ = 0.30. (d) n = 500, p̂ = 0.60. (e) n = 800, p̂ = 0.50.
Suppose that in a random sample of 450 employed Americans, there are 57 individuals who say that they would fire their boss if they could. Calculate a 90% confidence interval for the population proportion who would fire their boss if they could. (Round the answers to three decimal places.) to
In a randomly selected sample of 400 registered voters in a community, 140 individuals say that they plan to vote for Candidate Y in the upcoming election. (a) Find the sample proportion planning to vote for Candidate Y. (Round your answer to two decimal places.) (b) Calculate the standard error of the sample proportion. (Round your answer to three decimal places.) (c) Find a 95% confidence interval for the proportion of the registered voter population who plan to vote for Candidate Y. (Round your answers to three decimal places.) to (d) Find a 98% confidence interval for the proportion of the registered voter population who plan to vote for Candidate Y. (Round your answers to three decimal places.) to
In: Math
An urn contains 5 blue marbles and 4 yellow marbles. One marble is removed, its color noted, and not replaced. A second marble is removed and its color is noted.
(a) What is the probability that both marbles are blue? yellow?
(b) What is the probability that exactly one marble is blue?
A tree diagram has a root that splits into 2 branches labeled blue and yellow. Each primary branch splits into 2 secondary branches, labeled blue and yellow.yellowblueblueblueyellowyellow
In: Math
4. What is the empirical probability of a loss? [Topic 2]
Date OLIM Int.
15/6/2014 2.36
22/6/2014 2.46
29/6/2014 2.52
6/7/2014 2.46
13/7/2014 2.44
20/7/2014 2.54
27/7/2014 2.46
3/8/2014 2.42
10/8/2014 2.54
17/8/2014 2.53
24/8/2014 2.65
31/8/2014 2.64
7/9/2014 2.56
14/9/2014 2.54
21/9/2014 2.4
28/9/2014 2.3
5/10/2014 2.2
12/10/2014 2.08
19/10/2014 2.06
26/10/2014 2.13
2/11/2014 2.11
9/11/2014 2.25
16/11/2014 2.24
23/11/2014 2.16
30/11/2014 2.09
7/12/2014 2.04
14/12/2014 2.11
21/12/2014 2.09
28/12/2014 2.04
4/1/2015 2.01
11/1/2015 1.96
18/1/2015 2
25/1/2015 1.975
1/2/2015 2.03
8/2/2015 2
15/2/2015 2
22/2/2015 2
1/3/2015 2
8/3/2015 2.01
15/3/2015 1.98
22/3/2015 1.99
29/3/2015 2
5/4/2015 2.03
12/4/2015 2.05
19/4/2015 2
26/4/2015 2.02
3/5/2015 2
10/5/2015 1.98
17/5/2015 1.985
24/5/2015 1.985
31/5/2015 1.88
7/6/2015 1.885
14/6/2015 1.865
21/6/2015 1.865
28/6/2015 1.885
5/7/2015 1.825
12/7/2015 1.79
19/7/2015 1.78
26/7/2015 1.84
2/8/2015 1.8
9/8/2015 1.8
16/8/2015 1.755
23/8/2015 2.07
30/8/2015 1.98
6/9/2015 1.975
13/9/2015 2.04
20/9/2015 1.995
27/9/2015 2
4/10/2015 2
11/10/2015 2
18/10/2015 1.98
25/10/2015 2
1/11/2015 1.99
8/11/2015 1.915
15/11/2015 1.845
22/11/2015 1.82
29/11/2015 1.805
6/12/2015 1.77
13/12/2015 1.81
20/12/2015 1.835
27/12/2015 1.82
3/1/2016 1.695
10/1/2016 1.665
17/1/2016 1.63
24/1/2016 1.62
31/1/2016 1.61
7/2/2016 1.58
14/2/2016 1.585
21/2/2016 1.61
28/2/2016 1.755
6/3/2016 1.74
13/3/2016 1.745
20/3/2016 1.74
27/3/2016 1.69
3/4/2016 1.655
10/4/2016 1.72
17/4/2016 1.725
24/4/2016 1.65
1/5/2016 1.595
8/5/2016 1.6
15/5/2016 1.705
22/5/2016 1.815
29/5/2016 1.835
5/6/2016 1.86
12/6/2016 1.815
19/6/2016 1.855
26/6/2016 1.88
3/7/2016 1.91
10/7/2016 1.885
17/7/2016 1.88
24/7/2016 1.91
31/7/2016 1.83
7/8/2016 1.85
14/8/2016 1.96
21/8/2016 2.06
28/8/2016 2.07
4/9/2016 2.09
11/9/2016 2.03
18/9/2016 2.04
25/9/2016 2.06
2/10/2016 2.05
9/10/2016 2.07
16/10/2016 2.06
23/10/2016 2.1
30/10/2016 2.08
6/11/2016 2.1
13/11/2016 1.95
20/11/2016 1.96
27/11/2016 2.02
4/12/2016 2.07
11/12/2016 2.13
18/12/2016 2
25/12/2016 1.97
1/1/2017 2
8/1/2017 2.06
15/1/2017 1.995
22/1/2017 2
29/1/2017 2.01
5/2/2017 2.02
12/2/2017 2.1
19/2/2017 2.06
26/2/2017 2
5/3/2017 1.975
12/3/2017 1.93
19/3/2017 1.86
26/3/2017 1.92
2/4/2017 1.955
9/4/2017 1.91
16/4/2017 1.91
23/4/2017 1.91
30/4/2017 1.9
7/5/2017 1.96
14/5/2017 1.995
21/5/2017 2.07
28/5/2017 2.02
4/6/2017 2.03
11/6/2017 2
18/6/2017 1.96
25/6/2017 1.95
2/7/2017 1.915
9/7/2017 1.94
16/7/2017 1.945
23/7/2017 1.93
30/7/2017 1.96
6/8/2017 1.95
13/8/2017 2.02
20/8/2017 2.1
27/8/2017 2.06
3/9/2017 2.02
10/9/2017 2.01
17/9/2017 2.01
24/9/2017 2.02
1/10/2017 2.14
8/10/2017 2.22
15/10/2017 2.29
22/10/2017 2.35
29/10/2017 2.36
5/11/2017 2.33
12/11/2017 2.19
19/11/2017 2.2
26/11/2017 2.25
3/12/2017 2.19
10/12/2017 2.16
17/12/2017 2.07
24/12/2017 2.03
31/12/2017 2.04
7/1/2018 2.09
14/1/2018 2.11
21/1/2018 2.19
28/1/2018 2.22
4/2/2018 2.08
11/2/2018 2.17
18/2/2018 2.26
25/2/2018 2.23
4/3/2018 2.4
11/3/2018 2.34
18/3/2018 2.37
25/3/2018 2.34
1/4/2018 2.34
8/4/2018 2.35
15/4/2018 2.29
22/4/2018 2.28
29/4/2018 2.18
6/5/2018 2.3
13/5/2018 2.29
20/5/2018 2.28
27/5/2018 2.19
3/6/2018 2.21
10/6/2018 2.17
In: Math
A problem experiment is conducted in which the sample space of the experiment is S= {1,2,3,4,5,6,7,8,9,10,11,12}, event F={7,8}, and event G={9,10,11,12}. Assume that each outcome is equally likely. List the outcomes in F and G. Find P(F or G) by counting the number of outcomes in F or G. Determine P(F or G) using the general addition rule.
In: Math
Hormone replacement therapy (HRT) is thought to increase the risk of breast cancer. The accompanying data on
x = percent of women using HRT and
y = breast cancer incidence (cases per 100,000 women)
for a region in Germany for 5 years appeared in the paper "Decline in Breast Cancer Incidence after Decrease in Utilization of Hormone Replacement Therapy." The authors of the paper used a simple linear regression model to describe the relationship between HRT use and breast cancer incidence.
| HRT Use | Breast Cancer Incidence |
|---|---|
| 46.30 | 103.30 |
| 40.60 | 105.00 |
| 39.50 | 100.00 |
| 36.60 | 93.80 |
| 30.00 | 83.50 |
(a)
What is the equation of the estimated regression line? (Round your answers to three decimal places.)
ŷ = __+(___x)
(b)
What is the estimated average change in breast cancer incidence (in cases per 100,000 women) associated with a 1 percentage point increase in HRT use? (Round your answer to three decimal places.)
cases per 100,000 women
(c)
What would you predict the breast cancer incidence (in cases per 100,000 women) to be in a year when HRT use was 34%? (Round your answer to three decimal places.)
cases per 100,000 women
(d)
Should you use this regression model to predict breast cancer incidence for a year when HRT use was 13%? Explain.
(e)
Calculate the value of
r2.
(Round your answer to three decimal places.)Interpret the value of
r2.
(f)
Calculate the value of
se.
(Round your answer to three decimal places.)Interpret the value of
In: Math
|
The undergraduate grade point averages (UGPA) of students taking an admissions test in a recent year can be approximated by a normal distribution, as shown in the figure. (a) What is the minimum UGPA that would still place a student in the top55% of UGPAs?(b) Between what two values does the middle5050% of the UGPAs lie? |
3.3842.76Grade point
average
mu equals 3.38μ=3.38 sigma equals 0.19σ=0.19 x |
In: Math
A simple random sample of 33 men from a normally distributed population results in a standard deviation of 8.2 beats per minute. The normal range of pulse rates of adults is typically given as 60 to 100 beats per minute. If the range rule of thumb is applied to that normal range, the result is a standard deviation of 10 beats per minute. Use the sample results with a 0.10 significance level to test the claim that pulse rates of men have a standard deviation equal to 10 beats per minute. Complete parts (a) through (d) below.
a. Identify the null and alternative hypotheses.
b. Compute the test statistic; χ2 = ___ (Round to three decimal places as needed.)
c. Find the P-value; P-value = ____ (Round to four decimal places as needed.)
d. State the conclusion. (choose one from each ( x, y) set)
In: Math
In her book Red Ink Behaviors, Jean Hollands reports on the assessment of leading Silicon Valley companies regarding a manager's lost time due to inappropriate behavior of employees. Consider the following independent random variables. The first variable x1 measures manager's hours per week lost due to hot tempers, flaming e-mails, and general unproductive tensions. x1: 1 3 6 2 2 4 10 The variable x2 measures manager's hours per week lost due to disputes regarding technical workers' superior attitudes that their colleagues are "dumb and dispensable". x2: 8 3 2 7 9 4 10 3 (i) Use a calculator with sample mean and sample standard deviation keys to calculate x1, s1, x2, and s2. (Round your answers to two decimal places.) x1 = s1 = x2 = s2 = (ii) Does the information indicate that the population mean time lost due to hot tempers is different (either way) from population mean time lost due to disputes arising from technical workers' superior attitudes? Use α = 0.05. Assume that the two lost-time population distributions are mound-shaped and symmetric. (a) What is the level of significance? State the null and alternate hypotheses. H0: μ1 = μ2; H1: μ1 ≠ μ2 H0: μ1 = μ2; H1: μ1 < μ2 H0: μ1 = μ2; H1: μ1 > μ2 H0: μ1 ≠ μ2; H1: μ1 = μ2 (b) What sampling distribution will you use? What assumptions are you making? The standard normal. We assume that both population distributions are approximately normal with unknown standard deviations. The standard normal. We assume that both population distributions are approximately normal with known standard deviations. The Student's t. We assume that both population distributions are approximately normal with known standard deviations. The Student's t. We assume that both population distributions are approximately normal with unknown standard deviations. What is the value of the sample test statistic? (Test the difference μ1 − μ2. Do not use rounded values. Round your final answer to three decimal places.) (c) Find (or estimate) the P-value. P-value > 0.500 0.250 < P-value < 0.500 0.100 < P-value < 0.250 0.050 < P-value < 0.100 0.010 < P-value < 0.050 P-value < 0.010 Sketch the sampling distribution and show the area corresponding to the P-value. Maple Generated Plot Maple Generated Plot Maple Generated Plot Maple Generated Plot (d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis? Are the data statistically significant at level α? 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 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 statistically significant. At the α = 0.05 level, we reject the null hypothesis and conclude the data are not statistically significant. (e) Interpret your conclusion in the context of the application. Fail to reject the null hypothesis, there is sufficient evidence that there is a difference in mean time lost due to hot tempers and technical workers' attitudes. Fail to reject the null hypothesis, there is insufficient evidence that there is a difference in mean time lost due to hot tempers and technical workers' attitudes. Reject the null hypothesis, there is sufficient evidence that there is a difference in mean time lost due to hot tempers and technical workers' attitudes. Reject the null hypothesis, there is insufficient evidence that there is a difference in mean time lost due to hot tempers and technical workers' attitudes.
In: Math
A random sample of n1 = 10 regions in New England gave the following violent crime rates (per million population). x1: New England Crime Rate 3.5 3.7 4.2 3.9 3.3 4.1 1.8 4.8 2.9 3.1 Another random sample of n2 = 12 regions in the Rocky Mountain states gave the following violent crime rates (per million population). x2: Rocky Mountain Crime Rate 3.5 4.3 4.5 5.1 3.3 4.8 3.5 2.4 3.1 3.5 5.2 2.8 Assume that the crime rate distribution is approximately normal in both regions. (i) Use a calculator to calculate x1, s1, x2, and s2. (Round your answers to two decimal places.) x1 = s1 = x2 = s2 = (ii) Do the data indicate that the violent crime rate in the Rocky Mountain region is higher than in New England? Use α = 0.01. (a) What is the level of significance? State the null and alternate hypotheses. H0: μ1 = μ2; H1: μ1 < μ2 H0: μ1 = μ2; H1: μ1 ≠ μ2 H0: μ1 < μ2; H1: μ1 = μ2 H0: μ1 = μ2; H1: μ1 > μ2 (b) What sampling distribution will you use? What assumptions are you making? The Student's t. We assume that both population distributions are approximately normal with known standard deviations. The standard normal. We assume that both population distributions are approximately normal with unknown standard deviations. The standard normal. We assume that both population distributions are approximately normal with known standard deviations. The Student's t. We assume that both population distributions are approximately normal with unknown standard deviations. What is the value of the sample test statistic? (Test the difference μ1 − μ2. Round your answer to three decimal places.) (c) Find (or estimate) the P-value. P-value > 0.250 0.125 < P-value < 0.250 0.050 < P-value < 0.125 0.025 < P-value < 0.050 0.005 < P-value < 0.025 P-value < 0.005 Sketch the sampling distribution and show the area corresponding to the P-value. Maple Generated Plot Maple Generated Plot Maple Generated Plot Maple Generated Plot (d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis? Are the data statistically significant at level α? At the α = 0.01 level, we reject the null hypothesis and conclude the data are not statistically significant. At the α = 0.01 level, we fail to reject the null hypothesis and conclude the data are not statistically significant. At the α = 0.01 level, we reject the null hypothesis and conclude the data are statistically significant. At the α = 0.01 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. Reject the null hypothesis, there is sufficient evidence that violent crime in the Rocky Mountain region is higher than in New England. Fail to reject the null hypothesis, there is insufficient evidence that violent crime in the Rocky Mountain region is higher than in New England. Reject the null hypothesis, there is insufficient evidence that violent crime in the Rocky Mountain region is higher than in New England. Fail to reject the null hypothesis, there is sufficient evidence that violent crime in the Rocky Mountain region is higher than in New England.
In: Math
Please outline each step used along the way to solve the problem using excel only with cell numbers and formulas used. Thank you.
Whenever an Alliance Air customer flies on a prepurchased seat, Alliance Air obtains $100 in profits. However, if Alliance Air has more customers seeking a seat then they have prepurchased, Alliance Air is forced to book that passenger on a seat purchased that day. In such a situation, Alliance Air has a profit of -$170 due to the high cost of same-day flights. If not all of their prepurchased seats are taken, then Alliance Air makes a profit of -$20 by selling the seats at a discount to passengers outside of their customer base. Based on their data, Alliance Air knows that number of customers seeking a flight on any day follows a Poisson distribution with mean 40.
Using this information, complete the following tasks/questions:
| Alliance Air Service | ||||||||
| Yellow Cell is Input Cell | ||||||||
| Prepurchased Seat Sale | $ 100 | Avg Profit | Min Profit | Max Profit | % above 2250 | |||
| Same Day Flight | $ (170) | |||||||
| Discounted Flight Sale | $ (20) | |||||||
| Average Demand | 40 | |||||||
| Pre-Purchased Flights | 37 | |||||||
| Same Day Flights to purchase | ||||||||
| Discounted Flights Sold | ||||||||
| Daily Profit |
In: Math