1)
The
average weekly wage for a sample of 30 people employed in manufacturing is $580
with a standard deviation of $54. A sample of 40 people employed in a service
industry has an average weekly wage of $540 with a standard deviation of $48.
Based on these samples, is there any significant difference in the average
weekly wages of the 2 industries? Test at a 5% level of significance. (Z =
3.215; conclude significant difference in average weekly wages)
2)
Suppose
a level of significance had not been chosen. Why would the same conclusion be
reached? (p-value < 1%)
3)
Construct
a 95% confidence interval of the average difference in weekly wages between
manufacturing and service industry employees, rounding to the nearest cent.
Interpret the interval. If this interval were used to test the hypothesis in
question 1, why would the same conclusion be reached?
(15.61 < mean(manufacturing)
– mean(service) < 64.39)
4)
An
urban planner conducted a study in which a city was divided into different
regions. One of the regions was an old established neighbourhood;
another was a new subdivision. Annual household income statistics from the two
regions had the following results:
|
Old |
New |
|
|
Mean |
104,000 |
95,000 |
|
Standard
deviation |
15,400 |
12,600 |
|
Sample
size |
12 |
12 |
Analysis of the data indicates both samples are
normally distributed. Are the annual household incomes of those in the old neighbourhood significantly higher than those in the new
subdivision? Conduct all appropriate tests at a 5% level of significance. (t =
1.5669; conclude old neighbourhood is not
significantly higher)
5)
For
which levels of significance between 1% and 10% would the opposite conclusion
be reached? (6.57% and 10%)
6)
Construct
a 95% confidence interval of the average difference in average annual household
incomes between the old and new neighbourhoods,
rounding to the nearest hundred. If the average household income in the new neighbourhood is $95,000, in what range would the average
household income in the old neighbourhood lie?
(-$2,900 < mean(old) – mean(new) < $20,900)
7)
A
lab conducted an experiment on 13 lab rats to see if an experimental drug could
reduce the amount of time it would take the rats to go through a maze. The
results in seconds were:
|
Rat |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
10 |
11 |
12 |
13 |
|
Before |
113 |
94 |
99 |
77 |
81 |
91 |
111 |
104 |
85 |
66 |
111 |
51 |
109 |
|
After |
109 |
100 |
86 |
80 |
95 |
106 |
117 |
107 |
85 |
84 |
125 |
66 |
108 |
Analysis of the data indicates that the times
are normally distributed. Is the drug effective? Test at a 5% level of
significance. (t = -2.3117; do not reject Ho; conclude drug not effective)
8)
Suppose
a level of significance had not been chosen. Why would the same conclusion be
reached? (p-value > 10%)
9)
Construct
a 95% confidence interval of the average difference in times between before and
after the drug was administered, rounding to 1 decimal. Interpret the interval.
(-11.4 < md < -0.3)
10)
A
museum suspected that its members would be more likely to visit any new
exhibits than people from the general public. In the survey 200 from each group
were surveyed for a total of 400 respondents. Of the members who were surveyed,
170 said they would visit any new exhibits that year; the corresponding number
for respondents from the general public was 144. Are members from the museum
more likely to visit new exhibits? Test without assuming a level of
significance. (Z = 3.1644; conclude museum members more likely)
11)
Construct
a 99% confidence interval of the average difference in the percentage of
members and the general public who will visit new exhibits. Interpret the
interval.
(2.55% < p(members)
– p(public) < 23.45%)