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Two students have devised a dice game named “Sums” for their statistics class. The game consists of choosing to play odds or evens. Probabilities for “Sums”Roll23456789101112P(roll)Each person takes turns rolling two dice. If the sum is odd, the person playing odds gets points equal to the sum of the roll. If the sum is even, the person playing evens gets points equal to the sum of the roll. Note that the points earned is independent of who is rolling the dice.If Jessica is challenged to a game of Sums, which statement below is accurate in every aspect in guiding her to the correct choice of choosing to play odds or evens?

Question illustration
A
E(evens) will be more because there are more even numbers that result from rolling two dice. Therefore, Jessica should play evens.
B
E(odds) will be more because the probability for each odd number being rolled is greater. Therefore, Jessica should play odds.
C
E(evens) will be more because the value of the even numbers on the dice are more. Therefore, Jessica should play evens.
D
E(evens) = E(odds) because the different probabilities and values end up balancing out, creating a fair game. Therefore, Jessica may choose whichever she likes.
4

A simple random sample of size n is drawn from a normally distributed population, and the mean of the sample is , while the standard deviation is s. What is the 99% confidence interval for the population mean? Use the table below to help you answer the question.Confidence Level90%95%99%z*-score1.6451.962.58

Question illustration
A
x Overbar plus-or-minus StartFraction 0.90 times s Over StartRoot n EndRoot EndFraction
Option A
B
x Overbar plus-or-minus StartFraction 0.99 times s Over StartRoot n EndRoot EndFraction
Option B
C
x Overbar plus-or-minus StartFraction 1.645 times s Over StartRoot n EndRoot EndFraction
Option C
D
x Overbar plus-or-minus StartFraction 2.58 times s Over StartRoot n EndRoot EndFraction
Option D
5

The histogram below represents the number of points Mandy scored per game. The graph makes the data appear symmetric when it is actually skewed. How can the graph be adjusted to show the skew?

Question illustration
A
The intervals on the x-axis can be changed to be continuous and consistent.
B
The interval on the y-axis can be changed to show only the relevant numbers, 3 and 5.
C
The scale on the x-axis can be increased to show the interval 25–29.
D
The scale on the y-axis can be decreased to show only the numbers 0–5.
8

The mean of a set of credit scores is and . Which statement must be true about z694?

Question illustration
A
z694 is within 1 standard deviation of the mean.
B
z694 is between 1 and 2 standard deviations of the mean.
C
z694 is between 2 and 3 standard deviations of the mean.
D
z694 is more than 3 standard deviations of the mean.
9

When making a statistical inference about the mean of a normally distributed population based on a sample drawn from that population, which of the following statements is correct, all else being equal?

A
The 68% confidence interval is wider than the 90% confidence interval.
B
The 90% confidence interval is narrower than the 95% confidence interval.
C
The 90% confidence interval is wider than the 99% confidence interval.
D
The 99% confidence interval is narrower than the 68% confidence interval.
13

A convenience sample differs from a voluntary sample in that

A
a convenience sample is structured based on accessibility to the researcher, and a voluntary sample is based on participant interest.
B
convenience samples survey each participant once, and voluntary samples survey each participant numerous times.
C
convenience sampling is a method of random sampling, and a voluntary sample is not.
D
convenience sampling is not a probability-based method, and voluntary sampling is.
14

The table below shows the ages of houses to the nearest year in a neighborhood. Using the age of the houses as the random variable, X, which graph shows the probability distribution, PX(x), of a randomly chosen house?Age of HouseNumber of Houses115220325430

A
A probability distribution is shown. The probability of 1 is 0.16; 2 is 0.22; 3 is 0.28; 4 is 0.33.
Option A
B
A probability distribution is shown. The probability of 1 is 0.1; 2 is 0.2; 3 is 0.3; 4 is 0.4.
Option B
C
A probability distribution is shown. The probability of 1 is 0.65; 2 is 0.1; 3 is 0.12; 4 is 0.13.
Option C
D
A probability distribution is shown. The probability of 1 is 0.15; 2 is 0.2; 3 is 0.25; 4 is 0.3.
Option D
15

The stem-and-leaf plot below shows the number of pages each student in a class read the previous evening.Which statement is true about the data set?

Question illustration
A
Its median is greater than its mode.
B
It has a range of 52 pages.
C
The value of the first quartile is 13.
D
The data is symmetric.
17

For a standard normal distribution, which of the following expressions must always be equal to 1?

A
P (z less-than-or-equal-to a) minus P (negative a less-than-or-equal-to z less-than-or-equal-to z) minus P (z greater-than-or-equal-to a)
Option A
B
P (z less-than-or-equal-to a) minus P (negative a less-than-or-equal-to z less-than-or-equal-to z) + P (z greater-than-or-equal-to a)
Option B
C
P (z less-than-or-equal-to a) + P (negative a less-than-or-equal-to z less-than-or-equal-to z) minus P (z greater-than-or-equal-to a)
Option C
D
P (z less-than-or-equal-to a) + P (negative a less-than-or-equal-to z less-than-or-equal-to z) + P (z greater-than-or-equal-to a)
Option D
18

Emi computes the mean and variance for the population data set 87, 46, 90, 78, and 89. She finds the mean is 78. Her steps for finding the variance are shown below.What is the first error she made in computing the variance?

Question illustration
A
Emi failed to find the difference of 89 - 78 correctly.
B
Emi divided by N - 1 instead of N.
C
Emi evaluated (46 - 78)2 as -(32)2.
D
Emi forgot to take the square root of -135.6.

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