Question

1. The television program Nightline once asked viewers whether the United Nations should continue to have its headquarters in the United States. Two phone numbers weregiven. Viewers were asked to call one telephone number to respond “Yes” and the other to respond “No”. More than 186,000 callers responded and 67% said “No.” This isan example of

A. voluntary response sampling.
B. a survey with little bias because a large SRS was used.
C. a survey with little bias since someone who called would know his or her opinion.
D. all of the above.


2. The distribution of actual weights of 8 oz. wedges of cheddar cheese produced at a dairy is Normal with mean 8.1 ounces and standard deviation 0.1 ounces.
Reference: Ref 11-5

If a sample of five of these cheese wedges is selected, the probability that their average weight is less than 8 oz is (Use Table A)

A. 0.0125.
B. 0.1853.
C. 0.4871.
D. 0.9873.

3. A random variable can be described as

A. a probability of an event.
B. a variable whose value is a numerical outcome of a random phenomenon.
C. the proportion of times an event occurs over a long run of repeated trials.
D. all of the above.

4. A researcher plans to conduct a test of hypotheses at the ? = 0.05 significance level. She designs her study to have a power of 0.80 at a particular alternativevalue of the parameter of interest.

The probability that the researcher will commit a Type I error is
A. 0.05.
B. 0.20.
C. 0.80.
D. equal to the P-value and cannot be determined until the data have been collected.

5. At a large midwestern college, 4% of the students are Hispanic. A random sample of 20 students from the college are selected. Let X denote the number of Hispanicsamong them.

The mean of X is
A. 0.4.
B. 0.8.
C. 1.2.
D. 1.6.

6. Can brainwaves be used to measure job aptitude? During the 1980s, a psychologist tried to use brainwave measurements to identify U.S. Navy recruits who are good atusing a rifle (instead of simply watching them shoot). Several times each year, batches of data were analyzed. Each batch of data compared various brainwavemeasurements with rifle–shooting performance records for a group of recruits. The psychologist was looking for evidence of a relationship between brainwave measuresand shooting skill. During 1990, he analyzed 34 of these batches. In 32 of those batches, there was no evidence of association, but the other two batches werestatistically significant (at level of significance ? = .05) in demonstrating a relationship between brainwaves and shooting skill. Based on this, which of thefollowing is true?

A. The psychologist should toss out the 32 batches of data that failed to demonstrate an association between brainwaves and shooting skill, and keep the two batchesthat do demonstrate an association. After all, the grant that funds such research and pays the psychologist's salary needs to be renewed.
B. There's little evidence of an association. After all, since we're testing each batch of data at the ? = .05 level of significance, by chance alone 1 in 20 batcheswould demonstrate an association even if there really isn't one. This explains the two “good” batches the psychologist observed.
C. The psychologist should just pick one of the data sets (at random) and base his analysis on that.
D. All of the above.

7. Spelling mistakes in a text are either “nonword errors” or “word errors.” A nonword error produces a string of letters that is not a word, such as “the” typed as“teh.” Word errors produce the wrong word, such as “loose” typed as “lose.” Nonword errors make up 25% of all errors. A human proofreader will catch 90% of nonworderrors and 70% of word errors.

Of all the errors that the proofreader catches, what part are nonword errors?
A. 0.125
B. 0.225
C. 0.300
D. 0.375

Answer 1

Answer TO 2

the PDF of normal distribution is = 1/σ * √2π * e ^ -(x-u)^2/ 2σ^2

standard normal distribution is a normal distribution with a,

mean of 0,

standard deviation of 1

equation of the normal curve is ( Z )= x - u / sd/sqrt(n) ~ N(0,1)

mean ( u ) = 8.1

standard Deviation ( sd )= 0.1

sample size (n) = 4

P(X < 8) =   (8-8.1)/0.1/ Sqrt ( 4 )

= -0.1/0.05= -2

= P ( Z <-2) From Standard NOrmal Table

= 0.0228

the probability that their average weight is less than 8 oz is 0.0228

Answer 2

Option B explains it well:

OPTION B: There's little evidence of an association. After all, since we're testing each batch of data at the α = .05 level of significance, by chance alone 1 in 20 batches would demonstrate an association even if there really isn't one. This explains the two “good” batches the psychologist observed. [ANSWER, B]

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Note that alpha = 0.05 means the probability of a type I error, that is, rejecting a true Ho. So that explains why a few batches showed an association.