A study of identity theft looked at how well consumers protect themselves from this increasingly prevalent crime. The behaviors of 65 college students were compared with the behaviors of 56 nonstudents. One of the questions was "When asked to create a password, I have used either my mother's maiden name, or my pet's name, or my birth date, or the last four digits of my social security number, or a series of consecutive numbers." For the students, 24 agreed with this statement while 28 of the nonstudents agreed.
(a) Display the data in a two-way table.
Students | Nonstudents | Total | |||
Agreed | |||||
Disagreed | |||||
Total | 121 |
Perform the chi-square test. (Round your χ2 to
three decimal places and round your P-value to four
decimal places.)
χ2 | = | |
df | = | |
P-value | = |
Summarize the results.
We can conclude at the 5% level that students and nonstudents differ in the response to this question.We cannot conclude at the 5% level that students and nonstudents differ in the response to this question.
(b) Reanalyze the data using the methods for comparing two
proportions that we studied in the previous chapter. Compare the
results and verify that the chi-square statistic is the square of
the z statistic. (Test students who agreed minus
nonstudents who agreed. Round your z to two decimal places
and round your P-value to four decimal places.)
z | = | |
P-value | = |
(c) The students in this study were junior and senior college
students from two sections of a course in Internet marketing at a
large northeastern university. The nonstudents were a group of
individuals who were recruited to attend commercial focus groups on
the West Coast conducted by a lifestyle marketing organization.
Discuss how the method of selecting the subjects in this study
relates to the conclusions that can be drawn from it.
In: Statistics and Probability
3. A sample of 25 freshmen and senior students at a major university completed a survey that extent of satisfaction with the parking situation on the campus (the higher the score, the greater the satisfaction).
Freshmen |
11 43 30 30 45 41 12 40 42 35 45 25 10 33 50 42 32 38 11 47 22 26 37 38 10 |
Seniors |
10 45 35 28 52 35 12 50 40 30 40 10 10 38 12 40 15 30 20 43 23 25 30 40 12 |
Compute the median, mean and standard deviation scores for these data. Based on the measures of central tendency and dispersion, describe the change in the satisfaction with the parking service between freshman and seniors. Make sure that you make a reference to the possible shape of the underlying distributions for the two groups and the change in them over the period of four years? 5 Points
In: Statistics and Probability
Total plasma volume is important in determining the required plasma component in blood replacement therapy for a person undergoing surgery. Plasma volume is influenced by the overall health and physical activity of an individual. Suppose that a random sample of 42 male firefighters are tested and that they have a plasma volume sample mean of x = 37.5 ml/kg (milliliters plasma per kilogram body weight). Assume that σ = 7.20 ml/kg for the distribution of blood plasma.
(a) Find a 99% confidence interval for the population mean blood plasma volume in male firefighters. What is the margin of error? (Round your answers to two decimal places.)
lower limit
upper limit
margin of error
(b) What conditions are necessary for your calculations? (Select all that apply.)
-σ is unknown
-n is large
-the distribution of weights is uniform
-the distribution of weights is normal σ is known
(c) Interpret your results in the context of this problem.
-The probability that this interval contains the true average blood plasma volume in male firefighters is 0.99.
-1% of the intervals created using this method will contain the true average blood plasma volume in male firefighters.
-99% of the intervals created using this method will contain the true average blood plasma volume in male firefighters.
-The probability that this interval contains the true average blood plasma volume in male firefighters is 0.01.
(d) Find the sample size necessary for a 99% confidence level with maximal margin of error E = 2.30 for the mean plasma volume in male firefighters. (Round up to the nearest whole number.) male firefighters
In: Statistics and Probability
Assume that x has a normal distribution with the specified mean and standard deviation. Find the indicated probability. (Enter a number. Round your answer to four decimal places.)
μ = 22; σ = 3.4
P(x ≥ 30) =
Assume that x has a normal distribution with the specified mean and standard deviation. Find the indicated probability. (Enter a number. Round your answer to four decimal places.)
μ = 4; σ = 2
P(3 ≤ x ≤ 6) =
In: Statistics and Probability
Here is data with y as the response variable. x y 57.8 47.7 65.3 42.7 61 30.8 54.5 26.4 -70.7 -338.8 63 45.7 38.9 -46.1 70.2 35.5 Make a scatter plot of this data. Which point is an outlier? Enter as an ordered pair. For example (a,b) - with parenthesis. Find the regression equation for the data set without the outlier. Enter as an equation of the form y = a + b x . Rounded to three decimal places. Do not include the hat in y-hat. Find the regression equation for the data set with the outlier. Enter as an equation of the form y = a + b x . Rounded to three decimal places. Do not include the hat in y-hat. Is this outlier an influential point? No, the outlier does not appear to be an influential point. Yes, the outlier appears to be an influential point.
In: Statistics and Probability
Given P(A) = 0.6, P(B) = 0.5, P(A | B) = 0.3, do the following. (a) Compute P(A and B).
(b) Compute P(A or B).
In: Statistics and Probability
Can someone also tell me how to write the commands in excel for 1/T and ln(tgel). I have to plot 1/T versus ln(tgel).
Cure temp. | Gel time | Linearized Forms | ||
T (K) | tgel (h) | 1/T (K-1) | ln(tgel) | |
373.15 | 9.007 | |||
383.15 | 6.200 | |||
393.15 | 4.016 | |||
403.15 | 2.533 | |||
413.15 | 1.711 | |||
423.15 | 1.549 | |||
433.15 | 0.780 | |||
443.15 | 0.619 | |||
453.15 | 0.635 | |||
463.15 | 0.708 | |||
473.15 | 0.549 |
In: Statistics and Probability
Contrast frequent patterns, associations, and correlations.
In: Statistics and Probability
You are interested in finding a 95% confidence interval for the average commute that non-residential students have to their college. The data below show the number of commute miles for 12 randomly selected non-residential college students. Round answers to 3 decimal places where possible. 8 7 25 13 23 26 6 6 6 28 8 12 a. To compute the confidence interval use a distribution. b. With 95% confidence the population mean commute for non-residential college students is between and miles. c. If many groups of 12 randomly selected non-residential college students are surveyed, then a different confidence interval would be produced from each group. About percent of these confidence intervals will contain the true population mean number of commute miles and about percent will not contain the true population mean number of commute miles.
In: Statistics and Probability
Researchers often use z tests to compare their samples to known population norms. The Graded Naming Test (GNT) asks respondents to name objects in a set of 30 black-and-white drawings. The test, often used to detect brain damage, starts with easy words like kangaroo and gets progressively more difficult, ending with words like sextant. The GNT population norm for adults in England is 20.4. Roberts (2003) wondered whether a sample of Canadian adults had different scores than adults in England. If they were different, the English norms would not be valid for use in Canada. The mean for 30 Canadian adults was 17.5. For the purposes of this exercise, assume that the standard deviation of the adults in England is 3.2. Conduct all six steps of a z test. Step one: Identify the populations, Distribution, and assumptions. Step two: State the null and research hypothesis, in both words and symbolic notation. Step 3:Determine the characteristics of the comparison distribution. Step four: Determine the critical values, or cutt-offs, indicate the points beyond which we will reject the null hypothesis. Step five: Calculate the test statistic. Step six: Decide whether to reject or fail to reject the null hypothesis.
In: Statistics and Probability
It is well known that a placebo, a fake medication or treatment, can sometimes have a positive effect just because patients often expect the medication or treatment to be helpful. An article gave examples of a less familiar phenomenon, the tendency for patients informed of possible side effects to actually experience those side effects. The article cited a study in which a group of patients diagnosed with benign prostatic hyperplasia is randomly divided into two subgroups. One subgroup of size 55 received a compound of proven efficacy along with counseling that a potential side effect of the treatment is erectile dysfunction. The other subgroup of size 52 is given the same treatment without counseling. The percentage of the no-counseling subgroup that reported one or more sexual side effects is 19.23%, whereas 41.82% of the counseling subgroup reported at least one sexual side effect. State and test the appropriate hypotheses at significance level 0.05 to decide whether the nocebo effect is operating here. [Note: The estimated expected number of "successes" in the no-counseling sample is a bit shy of 10, but not by enough to be of great concern (some sources use a less conservative cutoff of 5 rather than 10).]
State the relevant hypotheses. (Use p1 for the true proportion of patients experiencing one or more sexual side effects when given no counseling and p2 for the true proportion of patients experiencing one or more sexual side effects when receiving counseling that a potential side effect of the treatment is erectile dysfunction.)
Calculate the test statistic and P-value. (Round your test statistic to two decimal places and your P-value to four decimal places.)
z= -2.54
p-value=?
In: Statistics and Probability
You are interested in finding a 90% confidence interval for the average number of days of class that college students miss each year. The data below show the number of missed days for 14 randomly selected college students. Round answers to 3 decimal places where possible. 1 8 10 9 1 3 1 8 1 0 2 1 5 8 a. To compute the confidence interval use a distribution. b. With 90% confidence the population mean number of days of class that college students miss is between and days. c. If many groups of 14 randomly selected non-residential college students are surveyed, then a different confidence interval would be produced from each group. About percent of these confidence intervals will contain the true population mean number of missed class days and about percent will not contain the true population mean number of missed class days.
In: Statistics and Probability
You are the Director of Purchasing for the WV Department of Education. Your agency has plans to privatize the meal service for 20 middle schools. Describe your bid specifications and how you would evaluate the bids.
In: Statistics and Probability
Computation of z-scores and standardized scores
You are a school psychologist and have been asked to administer a verbal intelligence test to a local senior high school English class to assess writing preparedness of the class. You have been given the mean (μ = 70) and population standard deviation (σ = 8) for the raw scores from this test. Based on the testing manual, you know that the developers of this assessment generate a standardized score using a predetermined mean of 100 and standard deviation of 25. Answer the following questions about the scores below.
In: Statistics and Probability
In: Statistics and Probability