ECON 203 Midterm 1 || with 100% Accurate Solutions.
Which of he following about the Chi-squared distribution is FALSE
a. We use the Chi-squared distribution to perform a test on variance.
b. The Chi-squared is not symmetrical.
c. Only positive Chi-squared values are possible.
d. The Chi-squared distribution is asymptotic to the horizontal axis to the right.
e. 1 is an approximation to the center of the distribution. correct answers e. 1 is an approximation
to the center of the distribution
REASON: The center of Chi-squared distribution is approximated by its degrees of freedom
minus one.
Assume that the distribution of final scores of golf players at the UC golf course is normal, with
a standard deviation of 3 and a median of 78. Consider that the par is 72, what percentage of
players score higher than the par? correct answers 97.5%
REASON: In a normal distribution, the mean equals the median. We are interested in finding the
percentage of players scoring higher than 72. Given that 72 is 2*σ to the left of the mean, we
know that the area to the left of 72 will be (1-.95)/2 = .025. The area to the right of 72 therefore
is 1-.025=.975 or 97.5%.
You have just performed a hypothesis test with a 5% level of significance. You have not rejected
the null hypothesis that the population mean is equal to 50 (H1: μ╪ 50). If you were to create a
90% confidence interval, what could you say about 50 being included or excluded from that
confidence interval?
a. The 90% confidence interval will include 50, since it is larger than the 95% confidence
interval.
b. The 90% confidence interval will include 50, since it is smaller than the 95% confidence
interval.
c. You are unsure if the 90% confidence interval includes 50, because the 95% confidence
interval contains 50.
d. The 90% confidence interval will not include 50, since it is smaller than the 95% confidence
interval.
e. You are unsure if the 90% confidence interval includes 50, because the 95% confidence
interval does not contain 50. correct answers C. You are unsure if the 90% confidence interval
includes 50, because the 95% confidence interval contains 50.
REASON: For the two-tailed test at the 5% level, we cannot reject the null; This means that 50
would be contained in a 95% confidence interval. However, since the 90% confidence interval is
smaller than the 95% CI, it is uncertain whether 50 will be contained or not in it.
Suppose you have a mound-shaped, symmetrical distribution, and now we add some extreme
values on the lower tail of the distribution. Which of the following could be true?
a. The distribution would be skewed to the right.
b. The distribution would be skewed to the left.
, c. The mode would be smaller than the median.
d. The mode would be smaller than the mean.
e. The mean would be larger than the median. correct answers B. The distribution would be
skewed to the left.
REASON: Starting from a mound-shaped, symmetrical distribution, adding some extreme values
on the lower tail (left side) would produce a fatter left tail. Therefore, the distribution would be
skewed to the left.
You decide to go to London as an exchange student for a year. When you arrive you realize that
everyone is crazy about soccer and you have to pick a winning local team to support. A friend
advises you to choose Arsenal over Chelsea since they win more games. Just to make sure his
advise is correct, you collect data on the number of games played and won by each team and find
that Arsenal (population 1) won 45 of the last 60 games, and Chelsea won 32 of the last 50
games.
What are the appropriate null and alternative hypotheses to test whether the proportion of games
won by Arsenal is higher than by Chelsea?
a. H0: p-hat1 - p-hat2 = 0 vs. p-hat1 - p-hat2 > 0
b. H0: p1 - p2 = 0 vs. H1: p1 - p2 < 0
c. H0: p1 - p2 =0 vs. H1: p1-p2> 0
d. H0: p1 - p2 > 0 vs. H1 : p1 - p2 = 0
e. H0: p-hat1 - p-hat2 = 0 vs. H1: p-hat1 - p-hat 2 ╪ 0 correct answers C. H0: p1 - p2 =0 vs. H1:
p1-p2> 0
REASON: Null and alternative hypotheses always deal in population parameters, so the ones
with p-hat are eliminated. You want to test whether the proportion of games won by Arsenal is
higher than that of Chelsea.
You decide to go to London as an exchange student for a year. When you arrive you realize that
everyone is crazy about soccer and you have to pick a winning local team to support. A friend
advises you to choose Arsenal over Chelsea since they win more games. Just to make sure his
advise is correct, you collect data on the number of games played and won by each team and find
that Arsenal (population 1) won 45 of the last 60 games, and Chelsea won 32 of the last 50
games.
What is the value of the appropriate test statistic?
a. 0.958
b. 0.987
c. 1.062
d. 1.254
e. 1.897 correct answers D. 1.254
REASON: We know that this is case 2 of difference in proportion since we are interested in
testing whether the proportion of wins for one team is larger than for the other. Our test statistic
will be z = [(p-hat1 - p-hat2)] / (SQRT(p-hat(1-p-hat)(1/n1+1/n2))
Which of he following about the Chi-squared distribution is FALSE
a. We use the Chi-squared distribution to perform a test on variance.
b. The Chi-squared is not symmetrical.
c. Only positive Chi-squared values are possible.
d. The Chi-squared distribution is asymptotic to the horizontal axis to the right.
e. 1 is an approximation to the center of the distribution. correct answers e. 1 is an approximation
to the center of the distribution
REASON: The center of Chi-squared distribution is approximated by its degrees of freedom
minus one.
Assume that the distribution of final scores of golf players at the UC golf course is normal, with
a standard deviation of 3 and a median of 78. Consider that the par is 72, what percentage of
players score higher than the par? correct answers 97.5%
REASON: In a normal distribution, the mean equals the median. We are interested in finding the
percentage of players scoring higher than 72. Given that 72 is 2*σ to the left of the mean, we
know that the area to the left of 72 will be (1-.95)/2 = .025. The area to the right of 72 therefore
is 1-.025=.975 or 97.5%.
You have just performed a hypothesis test with a 5% level of significance. You have not rejected
the null hypothesis that the population mean is equal to 50 (H1: μ╪ 50). If you were to create a
90% confidence interval, what could you say about 50 being included or excluded from that
confidence interval?
a. The 90% confidence interval will include 50, since it is larger than the 95% confidence
interval.
b. The 90% confidence interval will include 50, since it is smaller than the 95% confidence
interval.
c. You are unsure if the 90% confidence interval includes 50, because the 95% confidence
interval contains 50.
d. The 90% confidence interval will not include 50, since it is smaller than the 95% confidence
interval.
e. You are unsure if the 90% confidence interval includes 50, because the 95% confidence
interval does not contain 50. correct answers C. You are unsure if the 90% confidence interval
includes 50, because the 95% confidence interval contains 50.
REASON: For the two-tailed test at the 5% level, we cannot reject the null; This means that 50
would be contained in a 95% confidence interval. However, since the 90% confidence interval is
smaller than the 95% CI, it is uncertain whether 50 will be contained or not in it.
Suppose you have a mound-shaped, symmetrical distribution, and now we add some extreme
values on the lower tail of the distribution. Which of the following could be true?
a. The distribution would be skewed to the right.
b. The distribution would be skewed to the left.
, c. The mode would be smaller than the median.
d. The mode would be smaller than the mean.
e. The mean would be larger than the median. correct answers B. The distribution would be
skewed to the left.
REASON: Starting from a mound-shaped, symmetrical distribution, adding some extreme values
on the lower tail (left side) would produce a fatter left tail. Therefore, the distribution would be
skewed to the left.
You decide to go to London as an exchange student for a year. When you arrive you realize that
everyone is crazy about soccer and you have to pick a winning local team to support. A friend
advises you to choose Arsenal over Chelsea since they win more games. Just to make sure his
advise is correct, you collect data on the number of games played and won by each team and find
that Arsenal (population 1) won 45 of the last 60 games, and Chelsea won 32 of the last 50
games.
What are the appropriate null and alternative hypotheses to test whether the proportion of games
won by Arsenal is higher than by Chelsea?
a. H0: p-hat1 - p-hat2 = 0 vs. p-hat1 - p-hat2 > 0
b. H0: p1 - p2 = 0 vs. H1: p1 - p2 < 0
c. H0: p1 - p2 =0 vs. H1: p1-p2> 0
d. H0: p1 - p2 > 0 vs. H1 : p1 - p2 = 0
e. H0: p-hat1 - p-hat2 = 0 vs. H1: p-hat1 - p-hat 2 ╪ 0 correct answers C. H0: p1 - p2 =0 vs. H1:
p1-p2> 0
REASON: Null and alternative hypotheses always deal in population parameters, so the ones
with p-hat are eliminated. You want to test whether the proportion of games won by Arsenal is
higher than that of Chelsea.
You decide to go to London as an exchange student for a year. When you arrive you realize that
everyone is crazy about soccer and you have to pick a winning local team to support. A friend
advises you to choose Arsenal over Chelsea since they win more games. Just to make sure his
advise is correct, you collect data on the number of games played and won by each team and find
that Arsenal (population 1) won 45 of the last 60 games, and Chelsea won 32 of the last 50
games.
What is the value of the appropriate test statistic?
a. 0.958
b. 0.987
c. 1.062
d. 1.254
e. 1.897 correct answers D. 1.254
REASON: We know that this is case 2 of difference in proportion since we are interested in
testing whether the proportion of wins for one team is larger than for the other. Our test statistic
will be z = [(p-hat1 - p-hat2)] / (SQRT(p-hat(1-p-hat)(1/n1+1/n2))
