A simple random sample of 1100 males aged 12 to 17 in the United States were asked whether they played massive multiplayer online role-playing games (MMORPGs); 775 said that they did. a. We want to use this information to construct a 95% confidence interval to estimate the proportion of all U.S. males aged 12 to 17 who play MMORPGs.
State the parameter our confidence interval will estimate (in context).
b. State the value of our point estimate (i.e., the statistic, ). Round to four decimal places.
c. Identify the conditions that must be met to use this procedure and explain how you know that each one has been satisfied.
d. Find the appropriate critical value () and the standard error of the sample proportion (). SHOW YOUR WORK! Round the standard error to four decimal places.
e. Use the formula shown in your notes to get the 95% confidence interval by hand. SHOW YOUR WORK! Round to four decimal places.
f. Interpret the confidence interval constructed in part (e) in the context of the problem.
g. Interpret the confidence level in the context of the problem.
h. Suppose you wanted to estimate the proportion of 12-to-17 year-old males who play MMORPG’s with 95% confidence to have a margin of error within ± 2%. Calculate how large a sample you would need. Use the found in (b). SHOW YOUR WORK! Remember to round your final answer up to the nearest whole number.
i. If you wanted to have a margin of error of ±2% with 99% confidence, would your sample have to be larger, smaller, or the same size as the sample in part (h)? Explain.
j. Suppose MMORPG.com claims that 65% of all U.S. males aged 12-17 play massive multiplayer online role-playing games (MMORPGs). Does your 95% confidence interval support this claim?
k. What is the name of the significance test that can we perform to test the claim made?
l. What hypotheses would we use if we wished to conduct a two-sided test?
m. The only condition that changed from earlier is the Success/Failure Condition. You must now verify the expected number of successes and failures using our null hypothesis value.
n. Now we can proceed with the calculations of the standard deviation (). SHOW YOUR WORK! Round to four decimal places. Note: Remember to use and in the formula.
o. Calculate the z-score. SHOW YOUR WORK!
p. Use your z-score to calculate the p-value. SHOW YOUR WORK! Hint: Remember to double the p-value if you start by calculating a one-sided probability.
q. Interpret the p-value in context.
r. What decision would you draw based on the size of the p-value?
s. Are our confidence interval and significance test results in agreement?
In: Math
Approximately what percent of normally distributed data values lie within 2 standard deviation to either side of the mean?
In: Math
According to a publication, 12.3% of 18 to 25 dash year dash olds were users of marijuana in 2000. A recent poll of 1293 randomly selected 18 to 25 dash year dash olds revealed that 176 currently use marijuana. At the 5% significance level, do the data provide sufficient evidence to conclude that the percentage of 18 to 25 dash year dash olds who currently use marijuana has changed from the 2000 percentage of 12.3%? Use the one-proportion z-test to perform the appropriate hypothesis test, after checking the conditions for the procedure. What are the hypotheses for the one-proportion z-test? Upper H 0: pequals nothing; Upper H Subscript a: p ▼ greater than not equals less than nothing (Type integers or decimals.)
In: Math
In a study, researchers wanted to measure the effect of alcohol on the hippocampal region, the portion of the brain responsible for long-term memory storage, in adolescents. The researchers randomly selected 22 adolescents with alcohol use disorders to determine whether the hippocampal volumes in the alcoholic adolescents were less than the normal volume of 9.02cm3. An analysis of the sample data revealed that the hippocampal volume is approximately normal with
x overbarxequals=8.06cm3 and s=0.7cm3. Conduct the appropriate test at the α=0.01 level of significance.
(a) State the null and alternative hypothese
(b) Identify the t- statistic and p value
(c) Conclusion: (reject/Fail to reject) __ the null hypothesis. There (is/is not) __ sufficent evidence to claim that the mean hippocampal volume is (equal to/less than/greater than) __ __ cm3
In: Math
Salt (teaspoons) |
||||
Burner |
0 |
2 |
4 |
6 |
Right Back |
7(7) |
4(13) |
7(24) |
5(15) |
8(21) |
7(25) |
7(34) |
7(33) |
|
7(30) |
7(26) |
7(41) |
7(37) |
|
Right Front |
4(6) |
4(36) |
4(1) |
4(28) |
4(20) |
5(44) |
4(14) |
4(31) |
|
4(27) |
4(45) |
5(18) |
4(38) |
|
Left Back |
6(9) |
6(46) |
7(8) |
5(35) |
7(16) |
6(47) |
6(12) |
6(39) |
|
6(22) |
5(48) |
7(43) |
6(40) |
|
Left Front |
9(29) |
8(5) |
8(3) |
8(2) |
9(32) |
8(10) |
9(19) |
8(4) |
|
9(42) |
8(11) |
10(23) |
7(17) |
In: Math
The manager of the Groove Shopping Mall in Windhoek claims that visitors to this mall spend always more than 90 minutes in the mall on any occasion. to test this claim, the chairman of the Namibian chamber of commerce and Industry (NCCI) commissioned a study which found that, from a random sample of 25 recent visitors to this mall, the average visiting time was 95.5 minutes, with a standard deviation of 15 minutes. 2.3.1 Formulate a suitable null and alternative hypothesis for this situation 2.3.2 Compute the sample statistic 2.3.3 formulate the decision rule by using = 10% 2.3.4 What conclusion can the chairman of the Namibian Chamber of Commerce and Industry (NCCI) draw from the Findings?
In: Math
The Cotton Mill is an upscale chain of women's clothing stores, located in the southwestern United States. Do to recent success, The Cotton Mill's top management is planning to expand by locating new stores in other regions of the country. The director of planning has been asked to study the relationship between yearly sales and the store size. As part of the study, the director selects a sample of 25 stores and determines the size of the store in square feet and the sales for the last year. The sample data follows.
Store size (1000s of square feet) Sales ( millions of $ )
3.7 9.18
2.0 4.58
5.0 8.22
0.7 1.45
2.6 6.51
2.9 2.82
5.2 10.45
5.9 9.94
3.0 4.43
2.4 4.75
2.4 7.30
0.5 3.33
5.0 6.67
0.4 0.55
4.2 7.56
3.1 2.23
2.6 4.49
5.2 9.90
3.3 8.93
3.2 7.60
4.9 3.71
5.5 5.47
2.9 8.22
2.2 7.17
2.3 4.35
Using store size as the independent variable, run the data using excel and answer the following: 1. Write the regression equation. 2. Interpret the regression constant and regression coefficient. 3. Forecast a value for the dependent variable. 4. Test the significant of the regression coefficient using alpha = .05. 5. Test the overall significant of the regression model. 6. Interpret the coefficient of determination.
In: Math
Explain what it means conceptually when we find that a correlation — or any other statistic that is testing a hypothesis — is significant. Refer to type I error, alpha, and confidence level.
In: Math
A consumer research organization was interested in the influence of type of water on the effectiveness of a detergent. Test batches of washings were run in four randomly chosen machines having a particular type of water - soft, moderate, and hard. All batches had equal numbers of oil-stained rags and after each washing the number of rags stilled stained was determined. The following results were obtained.
Number of Rags with Stains
Sample
Observations Soft Moderate Hard
1 0 4 10
2 1 8 5
3 2 3 7
4 1 5 10
Using an alpha level of .01, would you conclude that the type of water influences the effectiveness of the detergent?
In: Math
Critics of television often refer to the detrimental effects that all the violence shown on television has on children. However, there may be another problem. It may be that watching television also reduces the amount of physical exercise, causing weight gains. A sample of 15 10-year-old children was taken. The number of pounds each child was overweight was recorded (a negative number indicates the child is underweight). In addition, the number of hours of television viewing per week was also recorded. These data are listed here. (also provided in the Excel Spreadsheet) Television 42 34 25 35 37 38 31 33 Overweight 18 6 0 −1 13 14 7 7 Television 19 29 38 28 29 36 18 Overweight −9 8 8 5 3 14 −7 REQUIRED a) Draw the scatter diagram. b) Calculate the sample regression line and describe what the coefficients tell you about the relationship between the two variables.
In: Math
You are a member of a group of researchers who study the reading ability of children. You would like to know if any of the following factors have an effect on the reading ability of a child: age, memory span, and IQ. You conduct a pilot study on a small group of 20 children. From the initial finding, you will make recommendations to your group on further research.
A. Build three different regression models the using the variables as shown below. Review the results and determine which is the model is the “best”.
Models |
Dependent variable |
Independent variables |
Model 1 |
reading_ability |
age |
Model 2 |
reading_ability |
memory_span |
Model 3 |
reading_ability |
iq |
1. What is the ANOVA table of the best model?
2. What is the regression equation of the best model?
3. Conduct the test for the significance of the Overall Regression Model for the best model?
4. What is R2 of the best model?
5. For the best model, what are the 95% confidence intervals for the estimates of the regression coefficients—the Bi’s?
6. Provide an interpretation of the slopes, bi’s.
age | memory_span | IQ | reading_ability | |
6.7 | 4.4 | 95 | 7.2 | |
5.9 | 4 | 90 | 6 | |
5.5 | 4.1 | 105 | 6 | |
6.2 | 4.8 | 98 | 6.6 | |
6.4 | 5 | 106 | 7 | |
7.3 | 5.5 | 100 | 7.2 | |
5.7 | 3.6 | 88 | 5.3 | |
6.15 | 5 | 95 | 6.4 | |
7.5 | 5.4 | 96 | 6.6 | |
6.9 | 5 | 104 | 7.3 | |
4.1 | 3.9 | 108 | 5 | |
5.5 | 4.2 | 90 | 5.8 | |
6.9 | 4.5 | 91 | 6.6 | |
7.2 | 5 | 92 | 6.8 | |
4 | 4.2 | 101 | 5.6 | |
7.3 | 5.5 | 100 | 7.2 | |
5.9 | 4 | 90 | 6 | |
5.5 | 4.2 | 90 | 5.8 | |
4 | 4.2 | 101 | 5.6 | |
5.9 | 4 | 90 | 6 | |
In: Math
Compare and contrast the deductive and inductive research models.
In: Math
The following table shows the Myers-Briggs personality preferences for a random sample of 406 people in the listed professions. E refers to extroverted and I refers to introverted.
Personality Type | |||
Occupation | E | I | Row Total |
Clergy (all denominations) | 67 | 40 | 107 |
M.D. | 73 | 89 | 162 |
Lawyer | 55 | 82 | 137 |
Column Total | 195 | 211 | 406 |
Use the chi-square test to determine if the listed occupations and personality preferences are independent at the 0.05 level of significance.
(a) What is the level of significance?
State the null and alternate hypotheses.
H0: Myers-Briggs preference and profession
are not independent
H1: Myers-Briggs preference and profession are
not independent.H0: Myers-Briggs preference and
profession are independent
H1: Myers-Briggs preference and profession are
not independent. H0:
Myers-Briggs preference and profession are not independent
H1: Myers-Briggs preference and profession are
independent.H0: Myers-Briggs preference and
profession are independent
H1: Myers-Briggs preference and profession are
independent.
(b) Find the value of the chi-square statistic for the sample.
(Round the expected frequencies to at least three decimal places.
Round the test statistic to three decimal places.)
Are all the expected frequencies greater than 5?
YesNo
What sampling distribution will you use?
Student's tuniform chi-squarenormalbinomial
What are the degrees of freedom?
(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 of independence?
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 Myers-Briggs preference and the profession are not independent.At the 5% level of significance, there is sufficient evidence to conclude that Myers-Briggs preference and the profession are not independent.
In: Math
The following table shows age distribution and location of a random sample of 166 buffalo in a national park.
Age | Lamar District | Nez Perce District | Firehole District | Row Total |
Calf | 15 | 13 | 13 | 41 |
Yearling | 12 | 8 | 13 | 33 |
Adult | 31 | 26 | 35 | 92 |
Column Total | 58 | 47 | 61 | 166 |
Use a chi-square test to determine if age distribution and location are independent at the 0.05 level of significance.
(a) What is the level of significance?
State the null and alternate hypotheses.
H0: Age distribution and location are not
independent.
H1: Age distribution and location are not
independent.H0: Age distribution and location
are independent.
H1: Age distribution and location are not
independent. H0: Age
distribution and location are not independent.
H1: Age distribution and location are
independent.H0: Age distribution and location
are independent.
H1: Age distribution and location are
independent.
(b) Find the value of the chi-square statistic for the sample.
(Round the expected frequencies to at least three decimal places.
Round the test statistic to three decimal places.)
Are all the expected frequencies greater than 5?
YesNo
What sampling distribution will you use?
Student's tbinomial normalchi-squareuniform
What are the degrees of freedom?
(c) Find or estimate the P-value of the sample test
statistic. (Round your answer to three decimal places.)
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 of independence?
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 sufficient evidence to conclude that age distribution and location are not independent.At the 5% level of significance, there is insufficient evidence to conclude that age distribution and location are not independent.
In: Math
The following table shows the Myers-Briggs personality preference and area of study for a random sample of 519 college students. In the table, IN refers to introvert, intuitive; EN refers to extrovert, intuitive; IS refers to introvert, sensing; and ES refers to extrovert, sensing.
Myers-Briggs Preference |
Arts & Science | Business | Allied Health | Row Total |
IN | 63 | 11 | 22 | 96 |
EN | 82 | 47 | 25 | 154 |
IS | 60 | 40 | 15 | 115 |
ES | 74 | 38 | 42 | 154 |
Column Total | 279 | 136 | 104 | 519 |
Use a chi-square test to determine if Myers-Briggs preference type is independent of area of study at the 0.05 level of significance.
(a) What is the level of significance?
State the null and alternate hypotheses.
H0: Myers-Briggs type and area of study are
independent.
H1: Myers-Briggs type and area of study are not
independent.H0: Myers-Briggs type and area of
study are not independent.
H1: Myers-Briggs type and area of study are
independent. H0:
Myers-Briggs type and area of study are not independent.
H1: Myers-Briggs type and area of study are not
independent.H0: Myers-Briggs type and area of
study are independent.
H1: Myers-Briggs type and area of study are
independent.
(b) Find the value of the chi-square statistic for the sample.
(Round the expected frequencies to at least three decimal places.
Round the test statistic to three decimal places.)
Are all the expected frequencies greater than 5?
YesNo
What sampling distribution will you use?
Student's tnormal chi-squarebinomialuniform
What are the degrees of freedom?
(c) Find or estimate the P-value of the sample test
statistic. (Round your answer to three decimal places.)
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 of independence?
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 sufficient evidence to conclude that Myers-Briggs type and area of study are not independent.At the 5% level of significance, there is insufficient evidence to conclude that Myers-Briggs type and area of study are not independent.
In: Math