SCORE
21/24
MAT
300
STATI
SUnit
5
Milest
one 5
Exam
UNIT 5 — MILESTONE 5
You passed this Milestone
21 questions were ans wered correctly.
2 questions were answered incorrectly.
1 question was skipped. These were marked incorrect.
1
Rachel recorded the number of call s she made at work during the week:
Day Calls
Monday 18
Tue sda y 14
Wedne sday 24
Thursday 16
She expected to make 18 calls each day. To answer whether the number of calls follows a uniform distribution, a chi-
square te st for goodne ss of fi t should be performed. (alpha = 0.10)
What i s the chi -squared te st stati sti c? Answers are rounded to the neare st hundredth.
3.27
3.11
4.61
1.52
RATI ONALE
Using the chi -square formula we can note the test statis tic is
CONCEPT
Chi-Square Test for Goodness-of-Fit
2
A re searcher ha s a table of data with 5 column variable s and 4 row variable s.
The value for the degree s of freedom in order to calculate the stati stic i s .
12
4
20
19
RATI ONALE
Recall to get the deg rees of freedom we use df = (r-1)(c-1) where c and r are the number of rows and columns. This
means df = (5-1)(4-1) = 4*3 =12.
CONCEPT
Chi-Square Test for Association and Independence
3
A market re search company conducted a survey of two groups of students from different school s. They found that
students from school A spent an average of 90 minute s studying daily, while the students from school B spent an
average of 75 minute s daily.
They want to find out if the difference in the mean times spent studying by the students of the two schools i s
statistically significant.
Which of the following sets show s the correct null hypothe si s and alternative hypothe si s?
Null H ypothesis: S chool B s tudents s pend more time s tudying than
School A.
Alternative Hypothesis: The difference in the mean times s pent by the
schools ' s tudents is 15 minutes.
Null H ypothesis: There is at least s ome difference in the mean times
spent by the s chools ' s tudents.
Alternative H ypothesis: The students from school B s pend more time
studying than the students from s chool A.
Null H ypothesis: T here is no difference in the mean times s pent by the
schools ' s tudents.
Alternative Hypothesis: There is at leas t s ome difference in the mean
times s pent by the s chools' s tudents.
Null H ypothesis: T he difference in the mean times s pent by the s chools'
students is 15 minutes.
Alternative Hypothesis: There is no difference in the mean times s pent
by the s chools' s tudents.
RATI ONALE
Recall that the null hypothesis is always of no difference.
So the null hypothesis (H o) is that the mean time studying for g roup A = mean for group B. T his would indicate no
difference between the two groups.
The alternative hypothesis (Ha) is that there is difference in the mean study time between the two groups.
CONCEPT
Hypothesis Testing
4
Joe i s measuring the widths of doors he bought to install in an apartment complex. He measured 72 doors a nd found a
mean width of 36.1 inches with a standard deviation of 0.3 inches. To test if the doors differ significantly from the
standard industry width of 36 inche s, he compute s a z-stati sti c.
What is the value of Joe's z-te st stati stic?
-1.81
-2.83
1.81
2.83
RATI ONALE
If we first note the denominator of
Then, g etting the z-s core we can note it is
This tells us that 36.1 is 2.83 standard deviations above the value of 36.
Note that when you round some values you may get s lightly different results, but the results should be relatively close
to this final calculated value.
CONCEPT
Z-Test for Population Means
5
Adam tabulated the value s for the average speeds on each day of hi s road trip a s 60. 5, 63.2, 54. 7, 51.6, 72. 3, 70.7, 67.2,
and 65. 4 mph. He wishe s to construct a 98% confi dence interval.
What value of t* should Adam use to construct the confidence interva l? Answer choice s are rounded to the
thousandths place.
2.896
2.517
2.998
4.489
RATI ONALE
Recall that we have n = 8, so the df = n-1 = 7. So if we go to the
row where df = 7 and then 0.01 for the tail probability, this giv es us a v alue of 2.998. R ecall that a 98% confidence
interv al would hav e 2% for the tails, s o 1% for each tail.
We can also us e the last row and find the corres ponding confidence lev el.
CONCEPT
How to Find a Critical T Value
6
For a left-tailed te st, the cri tical value of z so that a hypothe si s te st would reject the null hypothe si s at 1% signi ficance
level would be . Answer choice s are rounded to the hundredths place.
-3.09
-2.33
-1.03
-1.28
RATI ONALE
Recall that when a test statistic is smaller than in a left tailed
test we would reject Ho. If we go to the standard normal chart and use 1% or 0.01, we will search for the closest value to
1% as clos ely as possible.
0.0099 corres ponds with a z-s core of -2.33.
CONCEPT
How to Find a Critical Z Value
7
The government claim s tha t the average age of Californians i s 34 years. Joe hypothe size s that the average age of the
population of California is not equal to 34 years. He records a sam ple mean equal to 37 and state s the hypothe si s a s μ
= 34 vs μ ≠ 34.
What type of te st should Joe do?
Right-tailed test
Two-tailed test
Left-tailed test
Joe s hould not do any of the types of tes ts listed
RATI ONALE
Since the Ha is a not equal ( ) sign, this indicates he wants to run a two-tailed test where the rejection region is the
upper or lower tail.
CONCEPT
One-Tailed and Two-Tailed Tests
8
One condi tion for performi ng a hypothe si s te st i s that the observations are independent. Marta i s going to ta ke a
sa mple from a populati on of 600 students.
How many students will Marta have to sa mple without replacement to treat the observations a s independent?
60
540
120
300
RATI ONALE
In g eneral we want about 10% or less to s till assume independence.
So size = 0.1*N = 0.1(600) = 60
A sample of 60 or less would be s ufficient.
CONCEPT
Sampling With or Without Replacement
9
A table repre sents the number of students who pa ssed or failed an aptitude te st at two different campuse s.
South Campus North Campus
Passed 42 31
Failed 58 69
In order to determine if there i s a signi ficant difference between campuse s a nd pa ss ra te, the chi-square te st for
association and independence should be performed.
What i s the expected freque ncy of South Cam pus and pa ssed?
21/24
MAT
300
STATI
SUnit
5
Milest
one 5
Exam
UNIT 5 — MILESTONE 5
You passed this Milestone
21 questions were ans wered correctly.
2 questions were answered incorrectly.
1 question was skipped. These were marked incorrect.
1
Rachel recorded the number of call s she made at work during the week:
Day Calls
Monday 18
Tue sda y 14
Wedne sday 24
Thursday 16
She expected to make 18 calls each day. To answer whether the number of calls follows a uniform distribution, a chi-
square te st for goodne ss of fi t should be performed. (alpha = 0.10)
What i s the chi -squared te st stati sti c? Answers are rounded to the neare st hundredth.
3.27
3.11
4.61
1.52
RATI ONALE
Using the chi -square formula we can note the test statis tic is
CONCEPT
Chi-Square Test for Goodness-of-Fit
2
A re searcher ha s a table of data with 5 column variable s and 4 row variable s.
The value for the degree s of freedom in order to calculate the stati stic i s .
12
4
20
19
RATI ONALE
Recall to get the deg rees of freedom we use df = (r-1)(c-1) where c and r are the number of rows and columns. This
means df = (5-1)(4-1) = 4*3 =12.
CONCEPT
Chi-Square Test for Association and Independence
3
A market re search company conducted a survey of two groups of students from different school s. They found that
students from school A spent an average of 90 minute s studying daily, while the students from school B spent an
average of 75 minute s daily.
They want to find out if the difference in the mean times spent studying by the students of the two schools i s
statistically significant.
Which of the following sets show s the correct null hypothe si s and alternative hypothe si s?
Null H ypothesis: S chool B s tudents s pend more time s tudying than
School A.
Alternative Hypothesis: The difference in the mean times s pent by the
schools ' s tudents is 15 minutes.
Null H ypothesis: There is at least s ome difference in the mean times
spent by the s chools ' s tudents.
Alternative H ypothesis: The students from school B s pend more time
studying than the students from s chool A.
Null H ypothesis: T here is no difference in the mean times s pent by the
schools ' s tudents.
Alternative Hypothesis: There is at leas t s ome difference in the mean
times s pent by the s chools' s tudents.
Null H ypothesis: T he difference in the mean times s pent by the s chools'
students is 15 minutes.
Alternative Hypothesis: There is no difference in the mean times s pent
by the s chools' s tudents.
RATI ONALE
Recall that the null hypothesis is always of no difference.
So the null hypothesis (H o) is that the mean time studying for g roup A = mean for group B. T his would indicate no
difference between the two groups.
The alternative hypothesis (Ha) is that there is difference in the mean study time between the two groups.
CONCEPT
Hypothesis Testing
4
Joe i s measuring the widths of doors he bought to install in an apartment complex. He measured 72 doors a nd found a
mean width of 36.1 inches with a standard deviation of 0.3 inches. To test if the doors differ significantly from the
standard industry width of 36 inche s, he compute s a z-stati sti c.
What is the value of Joe's z-te st stati stic?
-1.81
-2.83
1.81
2.83
RATI ONALE
If we first note the denominator of
Then, g etting the z-s core we can note it is
This tells us that 36.1 is 2.83 standard deviations above the value of 36.
Note that when you round some values you may get s lightly different results, but the results should be relatively close
to this final calculated value.
CONCEPT
Z-Test for Population Means
5
Adam tabulated the value s for the average speeds on each day of hi s road trip a s 60. 5, 63.2, 54. 7, 51.6, 72. 3, 70.7, 67.2,
and 65. 4 mph. He wishe s to construct a 98% confi dence interval.
What value of t* should Adam use to construct the confidence interva l? Answer choice s are rounded to the
thousandths place.
2.896
2.517
2.998
4.489
RATI ONALE
Recall that we have n = 8, so the df = n-1 = 7. So if we go to the
row where df = 7 and then 0.01 for the tail probability, this giv es us a v alue of 2.998. R ecall that a 98% confidence
interv al would hav e 2% for the tails, s o 1% for each tail.
We can also us e the last row and find the corres ponding confidence lev el.
CONCEPT
How to Find a Critical T Value
6
For a left-tailed te st, the cri tical value of z so that a hypothe si s te st would reject the null hypothe si s at 1% signi ficance
level would be . Answer choice s are rounded to the hundredths place.
-3.09
-2.33
-1.03
-1.28
RATI ONALE
Recall that when a test statistic is smaller than in a left tailed
test we would reject Ho. If we go to the standard normal chart and use 1% or 0.01, we will search for the closest value to
1% as clos ely as possible.
0.0099 corres ponds with a z-s core of -2.33.
CONCEPT
How to Find a Critical Z Value
7
The government claim s tha t the average age of Californians i s 34 years. Joe hypothe size s that the average age of the
population of California is not equal to 34 years. He records a sam ple mean equal to 37 and state s the hypothe si s a s μ
= 34 vs μ ≠ 34.
What type of te st should Joe do?
Right-tailed test
Two-tailed test
Left-tailed test
Joe s hould not do any of the types of tes ts listed
RATI ONALE
Since the Ha is a not equal ( ) sign, this indicates he wants to run a two-tailed test where the rejection region is the
upper or lower tail.
CONCEPT
One-Tailed and Two-Tailed Tests
8
One condi tion for performi ng a hypothe si s te st i s that the observations are independent. Marta i s going to ta ke a
sa mple from a populati on of 600 students.
How many students will Marta have to sa mple without replacement to treat the observations a s independent?
60
540
120
300
RATI ONALE
In g eneral we want about 10% or less to s till assume independence.
So size = 0.1*N = 0.1(600) = 60
A sample of 60 or less would be s ufficient.
CONCEPT
Sampling With or Without Replacement
9
A table repre sents the number of students who pa ssed or failed an aptitude te st at two different campuse s.
South Campus North Campus
Passed 42 31
Failed 58 69
In order to determine if there i s a signi ficant difference between campuse s a nd pa ss ra te, the chi-square te st for
association and independence should be performed.
What i s the expected freque ncy of South Cam pus and pa ssed?
