Part I Review
1. a. The goal for descriptive statistics is to simplify, organize, and summarize
data so that it
is easier for researchers to see patterns.
b. A frequency distribution provides an organized summary of the complete set
of scores.
c. A measure of central tendency summarizes an entire set of scores with a
single value
that is representative of the whole set.
d. A measure of variability provides a single number that describes the
differences that
exist from one score to another.
2. a.
, The original rats appear to make far more errors that the seventh-generation
maze- bright
rats
b. The original rats made an average of M = 12.43 errors compared to an
average of only M
= 7.33 for the maze-bright rats. On average, the original rats made far more
errors.
c. For the original rats, SS = 427.14, the variance is s2 = 21.36 and the standard
deviation is
s = 4.62. For the maze-bright rats, SS = 54.67, the variance is s2 = 2.73
and the standard
deviation is s = 1.65. The error scores for the original rats are much more
spread out. The seventh generation rats are a much more homogeneous
group.
, Part II Review
1. a. z = 1.50
b. X = 36
c. If the entire population of X values is transformed into z-scores, the set of z-
score will
have a mean of 0 and a standard deviation of 1.00.
d. The standard error is 4 points and z = 0.50.
e. The standard error is 2 points and z = 1.00.
2. a. p(X > 40) = p(z > 0.36) = 0.3594 or 35.94%.
b. p(X < 10) = p(z < –1.79) = 0.0367 or 3.67%.
c. The standard error is 2 points and z = –2.50. The probability is p = 0.0062.
3. a. The null hypothesis states that the overweight students are no different from
the overall
population, μ = 4.22. The standard error is 0.10 and the z-score for this
sample is z =
2.60. Reject the null hypothesis. The number of snacks eaten by
overweight students is significantly different from the number for the
general population.
b. The null hypothesis states that the healthy-weight students do not eat fewer
snacks than the overall population, H0: μ ≥ 4.22. The standard error is 0.12
and the z-score
, for this sample is z = –1.75. For a one-tailed test, the critical value is z = –
1.65. Reject
the null hypothesis. The number of snacks eaten by healthy-weight students
is
significantly less than the number for the general population.
1. a. The goal for descriptive statistics is to simplify, organize, and summarize
data so that it
is easier for researchers to see patterns.
b. A frequency distribution provides an organized summary of the complete set
of scores.
c. A measure of central tendency summarizes an entire set of scores with a
single value
that is representative of the whole set.
d. A measure of variability provides a single number that describes the
differences that
exist from one score to another.
2. a.
, The original rats appear to make far more errors that the seventh-generation
maze- bright
rats
b. The original rats made an average of M = 12.43 errors compared to an
average of only M
= 7.33 for the maze-bright rats. On average, the original rats made far more
errors.
c. For the original rats, SS = 427.14, the variance is s2 = 21.36 and the standard
deviation is
s = 4.62. For the maze-bright rats, SS = 54.67, the variance is s2 = 2.73
and the standard
deviation is s = 1.65. The error scores for the original rats are much more
spread out. The seventh generation rats are a much more homogeneous
group.
, Part II Review
1. a. z = 1.50
b. X = 36
c. If the entire population of X values is transformed into z-scores, the set of z-
score will
have a mean of 0 and a standard deviation of 1.00.
d. The standard error is 4 points and z = 0.50.
e. The standard error is 2 points and z = 1.00.
2. a. p(X > 40) = p(z > 0.36) = 0.3594 or 35.94%.
b. p(X < 10) = p(z < –1.79) = 0.0367 or 3.67%.
c. The standard error is 2 points and z = –2.50. The probability is p = 0.0062.
3. a. The null hypothesis states that the overweight students are no different from
the overall
population, μ = 4.22. The standard error is 0.10 and the z-score for this
sample is z =
2.60. Reject the null hypothesis. The number of snacks eaten by
overweight students is significantly different from the number for the
general population.
b. The null hypothesis states that the healthy-weight students do not eat fewer
snacks than the overall population, H0: μ ≥ 4.22. The standard error is 0.12
and the z-score
, for this sample is z = –1.75. For a one-tailed test, the critical value is z = –
1.65. Reject
the null hypothesis. The number of snacks eaten by healthy-weight students
is
significantly less than the number for the general population.
