A transect is an archaeological study area that is 1/5 mile wide and 1 mile long. A site in a transect is the location of a significant archaeological find. Let x represent the number of sites per transect. In a section of Chaco Canyon, a large number of transects showed that x has a population variance σ2 = 42.3. In a different section of Chaco Canyon, a random sample of 20 transects gave a sample variance s2 = 46.5 for the number of sites per transect. Use a 5% level of significance to test the claim that the variance in the new section is greater than 42.3. Find a 95% confidence interval for the population variance.
(a) What is the level of significance?
State the null and alternate hypotheses.
Ho: σ2 = 42.3; H1: σ2 ≠ 42.3 Ho: σ2 = 42.3; H1: σ2 > 42.3 Ho: σ2 = 42.3; H1: σ2 < 42.3 Ho: σ2 > 42.3; H1: σ2 = 42.3
(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 exponential population distribution. We assume a normal population distribution. We assume a binomial population distribution. We assume a uniform population distribution.
(c) Find or estimate the P-value of the sample test
statistic.
P-value > 0.100 0.050 < P-value < 0.100 0.025 < P-value < 0.050 0.010 < P-value < 0.025 0.005 < P-value < 0.010 P-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 conclude that the variance is greater in the new section. At the 5% level of significance, there is sufficient evidence to conclude conclude that the variance is greater in the new section.
(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 95% confident that σ2 lies outside this interval. We are 95% confident that σ2 lies above this interval. We are 95% confident that σ2 lies below this interval. We are 95% confident that σ2 lies within this interval.
In: Math
Discuss validity and reliability as "confounding variables" and how can they negatively impact the results of a study? What are confounding variables? Try to provide an example to illustrate your point.
In: Math
A poll surveyed people in six countries to assess attitudes toward a variety of alternate forms of energy. Suppose the data in the following table are a portion of the poll's findings concerning whether people favor or oppose the building of new nuclear power plants.
Response | Country | |||||
---|---|---|---|---|---|---|
Great Britain |
France | Italy | Spain | Germany | United States |
|
Strongly favor | 298 | 161 | 141 | 128 | 133 | 204 |
Favor more than oppose | 309 | 367 | 348 | 272 | 222 | 326 |
Oppose more than favor | 219 | 334 | 381 | 322 | 311 | 316 |
Strongly oppose | 220 | 215 | 217 | 389 | 443 | 174 |
(a)
How large was the sample in this poll?
(b)
Conduct a hypothesis test to determine whether people's attitude toward building new nuclear power plants is independent of country.
State the null and alternative hypotheses.
H0: The attitude toward building new nuclear
power plants is not mutually exclusive of the country.
Ha: The attitude toward building new nuclear
power plants is mutually exclusive of the
country.H0: The attitude toward building new
nuclear power plants is not independent of the country.
Ha: The attitude toward building new nuclear
power plants is independent of the
country. H0: The
attitude toward building new nuclear power plants is mutually
exclusive of the country.
Ha: The attitude toward building new nuclear
power plants is not mutually exclusive of the
country.H0: The attitude toward building new
nuclear power plants is independent of the country.
Ha: The attitude toward building new nuclear
power plants is not independent of the country.
Find the value of the test statistic. (Round your answer to two decimal places.)
Find the p-value. (Round your answer to four decimal places.)
p-value =
State your conclusion.
Do not reject H0. We cannot conclude that the attitude toward building new nuclear power plants is independent of the country.Reject H0. We conclude that the attitude toward building new nuclear power plants is not independent of the country. Reject H0. We cannot conclude that the attitude toward building new nuclear power plants is independent of the country.Do not reject H0. We conclude that the attitude toward building new nuclear power plants is not independent of the country.
(c)
Using the percentage of respondents who "strongly favor" and "favor more than oppose," which country has the most favorable attitude toward building new nuclear power plants?
Great BritainFrance ItalySpainGermanyUnited States
Which country has the least favorable attitude?
Great BritainFrance ItalySpainGermanyUnited States
In: Math
Use Excel to test. For each paired difference, compute After – Before. In Data Analysis, t-Test: Paired Two Sample for means, select the After data for Variable 1 Range. Note that the critical value output by Data Analysis for this test is always positive. In this problem, the sign of the critical value is negative corresponding to 1-tailed test with lower reject region and negative lower critical value.
Person Before After
1 176 164
2 192 191
3 185 176
4 177 176
5 196 185
6 178 169
7 196 196
8 181 172
9 158 158
10 201 193
11 191 185
12 193 189
13 176 175
14 212 210
15 177 173
16 183 180
17 210 204
18 198 192
19 157 152
20 213 200
21 161 161
22 177 166
23 210 203
24 192 186
25 178 170
What is your conclusion?
A. Do not reject the NULL Hypothesis because the actual value is greater than critical value
B. Reject the NULL Hypothesis because the actual value is less than the critical value
C. Reject the NULL Hypothesis because the actual value is greater than critical value
D. Do not reject the NULL Hypothesis because the actual value is less than the critical value
In: Math
11. Assume that when adults with smartphones are randomly selected, 57% use them in meetings or classes. If 20 adult smartphone users are randomly selected, find the probability that exactly 15 of them use their smartphones in meetings or classes.
The probability is _____
(Round to four decimal places as needed.)
12. Assume that when adults with smartphones are randomly selected, 58% use them in meetings or classes. If 10 adult smartphone users are randomly selected, find the probability that at least 7 of them use their smartphones in meetings or classes.
The probability is_____
(Round to four decimal places as needed.)
13. A survey showed that 76% of adults need correction (eyeglasses, contacts, surgery, etc.) for their eyesight. If 8 adults are randomly selected, find the probability that no more than 1 of them need correction for their eyesight. Is 1 a significantly low number of adults requiring eyesight correction?
The probability that no more than 1 of the 8 adults require eyesight correction is _____.
(Round to three decimal places as needed.)
In: Math
Perform t-test on eRPM for Strategy A and B. H0: A = B vs. H1: A != B (Two-sided t-test) What is the p-value?
Strategy | Date | eRPM |
A | 1-Jun-14 | 3.33 |
A | 2-Jun-14 | 2.94 |
A | 3-Jun-14 | 3.03 |
A | 4-Jun-14 | 2.99 |
A | 5-Jun-14 | 3.08 |
A | 6-Jun-14 | 3.14 |
A | 7-Jun-14 | 3.32 |
A | 8-Jun-14 | 3.27 |
A | 9-Jun-14 | 3.15 |
A | 10-Jun-14 | 3.24 |
A | 11-Jun-14 | 3.2 |
A | 12-Jun-14 | 3.21 |
A | 13-Jun-14 | 3.25 |
A | 14-Jun-14 | 3.48 |
A | 15-Jun-14 | 3.47 |
A | 16-Jun-14 | 3.25 |
A | 17-Jun-14 | 3.32 |
A | 18-Jun-14 | 3.46 |
A | 19-Jun-14 | 3.58 |
A | 20-Jun-14 | 3.48 |
A | 21-Jun-14 | 3.48 |
A | 22-Jun-14 | 3.46 |
A | 23-Jun-14 | 3.34 |
A | 24-Jun-14 | 3.33 |
A | 25-Jun-14 | 3.37 |
A | 26-Jun-14 | 3.53 |
A | 27-Jun-14 | 3.67 |
A | 28-Jun-14 | 3.83 |
A | 29-Jun-14 | 3.78 |
A | 30-Jun-14 | 3.48 |
B | 1-Jun-14 | 2.95 |
B | 2-Jun-14 | 2.59 |
B | 3-Jun-14 | 2.76 |
B | 4-Jun-14 | 3 |
B | 5-Jun-14 | 3.24 |
B | 6-Jun-14 | 3.43 |
B | 7-Jun-14 | 3.44 |
B | 8-Jun-14 | 3.46 |
B | 9-Jun-14 | 3.27 |
B | 10-Jun-14 | 3.39 |
B | 11-Jun-14 | 3.37 |
B | 12-Jun-14 | 3.32 |
B | 13-Jun-14 | 3.49 |
B | 14-Jun-14 | 3.53 |
B | 15-Jun-14 | 3.34 |
B | 16-Jun-14 | 3.3 |
B | 17-Jun-14 | 3.33 |
B | 18-Jun-14 | 3.6 |
B | 19-Jun-14 | 3.85 |
B | 20-Jun-14 | 3.89 |
B | 21-Jun-14 | 3.69 |
B | 22-Jun-14 | 3.64 |
B | 23-Jun-14 | 3.6 |
B | 24-Jun-14 | 3.42 |
B | 25-Jun-14 | 3.41 |
B | 26-Jun-14 | 3.72 |
B | 27-Jun-14 | 3.94 |
B | 28-Jun-14 | 4.07 |
B | 29-Jun-14 | 4.05 |
B | 30-Jun-14 | 3.69 |
In: Math
Choose an organization (Amazon, Sony, etc.) and discuss how they manage large data sets including protocol for transferring data. Select one public data set and examine the technical format and how the data is manipulated globally.
In: Math
You have 100 coins, and 99 of them are fair (equal probability of heads or tails). One of them is weighted and has a 90% probability of landing on heads. You randomly choose one of the 100 coins. Find the probability that it is a weighted coin, under the following scenarios: (Hint: if your calculator can’t compute 100!, R can, just type factorial(100))
(a) You flip it 10 times and lands on heads 10 times (b) You flip it 10 times and it lands on heads 9 times
(c) You flip it 20 times and it lands on heads 18 times (d) You flip it 100 times and it lands on heads 77 times
In: Math
Consider the following data set.
x | 1 | 2 | 3 | 4 | 5 | 6 |
y | 3.00 | 0.21 | 0.61 | 0.70 | 1.13 | 1.17 |
a) plot the data (y versus x). Are there any points that appear to be outliers? If there are, circle them and label as such.
b) produce a regression of y against x. Add the regression line to the plot in a). Do you think that the regression line captures the most important features of the data set reasonably well?
c) using calculations at a 5% significance level, can you say that there is a significant linear relationship between the x and y? That is, can you say with 95% confidence that y linearly depends on x? Does this result agree with the conclusion you made in b)?
d) testing at a 5% significance level, can you say that the intercept (β0) is not zero? How does this conclusion agree with the plot in b)?
e) Assume that the first data point is an outlier (e.g. the value was misrecorded). Remove the outlier, and redo the parts b)-d). Plot the data set and both regression lines (before and after the outlier was removed). Comment on the difference. Also comment on the difference between the results of the tests in c) and d), if any.
In: Math
Perform t-test on eRPM for strategy A and B. H0: A = B vs. H1: A != B (Two-sided t-test) What is the p-value?
Strategy | Date | eRPM |
A | 15-Jun-14 | 3.47 |
A | 16-Jun-14 | 3.25 |
A | 17-Jun-14 | 3.32 |
A | 18-Jun-14 | 3.46 |
A | 19-Jun-14 | 3.58 |
A | 20-Jun-14 | 3.48 |
A | 21-Jun-14 | 3.48 |
A | 22-Jun-14 | 3.46 |
A | 23-Jun-14 | 3.34 |
A | 24-Jun-14 | 3.33 |
A | 25-Jun-14 | 3.37 |
A | 26-Jun-14 | 3.53 |
A | 27-Jun-14 | 3.67 |
A | 28-Jun-14 | 3.83 |
A | 29-Jun-14 | 3.78 |
A | 30-Jun-14 | 3.48 |
B | 15-Jun-14 | 3.34 |
B | 16-Jun-14 | 3.3 |
B | 17-Jun-14 | 3.33 |
B | 18-Jun-14 | 3.6 |
B | 19-Jun-14 | 3.85 |
B | 20-Jun-14 | 3.89 |
B | 21-Jun-14 | 3.69 |
B | 22-Jun-14 | 3.64 |
B | 23-Jun-14 | 3.6 |
B | 24-Jun-14 | 3.42 |
B | 25-Jun-14 | 3.41 |
B | 26-Jun-14 | 3.72 |
B | 27-Jun-14 | 3.94 |
B | 28-Jun-14 | 4.07 |
B | 29-Jun-14 | 4.05 |
B | 30-Jun-14 | 3.69 |
In: Math
#2. The operations manager of a musical instrument distributor feels that demand for a particular type of guitar may be related to the number of YouTube views for a popular music video by the popular rock group Marble Pumpkins during the preceding month. The manager has collected the data shown in the following table: YouTube Views (1000s) Guitar Sales 30 8 40 11 70 12 60 10 80 15 50 13
a. Graph the data to see whether a linear equation might describe the relationship between the views on YouTube and guitar sales.
b. Using the equations presented in this chapter, compute the SST, SSE, and SSR. Find the least squares regression line for the data.
c. Using the regression equation, predict guitar sales if there were 40,000 views last month.
In: Math
(1) The table below is a probability distribution of potential quantity of sales of Gourmet sausages during a game. John Bull has to pay a concession fee of $200 to receive a permit to sell sausages at the stadium. Gourmet sausages can be bought at wholesale for $2.00 and sold in the stadium for $3.50 each. Unsold sausages cannot be returned. Given the probability distribution:
SALES |
Probability |
100 |
0.05 |
150 |
0.06 |
200 |
0.10 |
250 |
0.20 |
300 |
0.25 |
350 |
0.30 |
400 |
0.04 |
In: Math
1. The president of a national real estate company wanted to know why certain branches of the company outperformed others. He felt that the key factors in determining total annual sales ($ in millions) were the advertising budget (in $1000s) X1 and the number of sales agents X2. To analyze the situation, he took a sample of 25 offices and ran the following regression. The computer output is below.
PREDICTOR COEF STDEV P-VALUE
Constant -19.47 15.84 0.2422
X1 0.1584 .0561 0.0154
X2 0.9625 .7781 0.2386
Se = 7.362 R squared = .524 Sig F = 0.0116
(a)What are the anticipated signs for each of the independent variables in the model?
(b) Interpret the slope coefficient associated with the number of real estate agents.
(c) Test to determine if a positive relationship exists between the advertising spending and annual sales. Use alpha = .05.
(d) Can we conclude that this model explains a significant portion of the variation in annual sales? Use alpha =.01.
In: Math
graduate student wants to estimate the number of research participants he will see in the fall semester. Using his data from the previous nine semesters, he tabulates a mean of 140 students per semester, although departmental records reflect a seasonal variation (i.e., population standard deviation) of 45 students. Calculate the 99% confidence interval.
In: Math
In: Math