Title: Basic Quantitative Methods in the Social Sciences (AKA Intro Stats)
1Basic Quantitative Methods in the Social
Sciences(AKA Intro Stats)
- 02-250-01
- Lecture 11 (Review Revised June 18 1015 PM)
2A Few Things
- You need to be able to decide which hypothesis
test statistic to use for the exam questions. We
have covered (since the last midterm) - z-test
- One-sample t-test
- Two sample t-tests
- Independent Sample t
- Dependent (Related) Samples t
3A Few Hints
- z-test
- Comparing a sample mean with a population mean
- Sigma is known (so you calculate standard error)
- One-sample t-test
- Comparing a sample mean with a population mean
- Sigma is unknown (so you are given or need to
calculate sample standard deviation to get
estimated standard error)
4A Few Hints
- Independent Sample t
- Comparing two sample means
- Two distinct (independent) groups
- Dependent (Related) Samples t
- Comparing two sample means
- Scores are pairs (e.g., pre-post test, couples,
members of the same family, etc)
5Question 1
- Researcher Q knows that regular Canadians spend
an average of 50 per month on take-out food. She
thinks that students probably spend more money on
take out-food per month than regular Canadians.
She asks 25 students how much money they spend on
take-out food per month, and finds that they
spend an average of 75 per month, with a
standard deviation of 10.
6Question 1
- A) State the null and alternative hypotheses
- B) What type of test should you use?
- C) Test the researchers hypothesis at the .01
level of significance - D) Test the researchers hypothesis at the .05
level of significance using the Confidence
Interval approach
7Question 1
- A) State the null and alternative hypotheses
- Ho students spend the same amount of money on
take-out food per month as regular Canadians - Ha students spend more money on take-out food
per month than regular Canadians - B) What type of test should you use?
- We are comparing a sample mean with a population
mean, but do not know sigma - We use a ONE-SAMPLE T-TEST
8Question 1
- C) Test the researchers hypothesis at the .01
level of significance - Alpha .01 (one-tailed), df 24, tcrit 2.492
- B/c tobs gt tcrit, we reject the Ho
- Conclusion students spend more money on take-out
food each month than regular Canadians
9- D) Test the researchers hypothesis at the .05
level of significance using the Confidence
Interval approach - B/c is outside of the interval, we
reject Ho - Conclusion students spend more money on take-out
food each month than regular Canadians
10Question 2
- Dr. Smith thinks that pet owners are happier than
people who do not own pets. She randomly samples
5 dog owners and 6 people who do not own pets and
asks them to rate their level of happiness on the
Happy Scale (from 1 to 50). The pet owners have
a mean Happy Scale rating of 35 with a standard
deviation of 5, while the people who do not own
pets have a mean Happy Scale rating of 30 with
a standard deviation of 3.
11Question 2
- A) State the null and alternative hypotheses
- B) State the IV, levels of IV, and DV
- C) What type of test should you use?
- D) Test Dr. Smiths hypothesis at the .05 level
of significance - E) Test Dr. Smiths hypothesis at the .05 level
of significance using the Confidence Interval
approach
12- A) State the null and alternative hypotheses
- Ho people who own pets are as happy as people
who do not own pets - Ha people who own pets are happier than people
who do not own pets - B) State the IV, levels of IV, and DV
- IV pet ownership (yes, no)
- DV Happy Scale rating (happiness)
- C) What type of test should you use?
- Independent Samples t-test
13- D) Test Dr. Smiths hypothesis at the .05 level
of significance
tcrit 1.833 lt tobs, reject Ho Conclusion pet
owners are happier than those who do not own pets
df (n1 n2) 2 11 - 2 9
14- E) Test Dr. Smiths hypothesis at the .05 level
of significance using the Confidence Interval
approach - B/c 0 is within the interval, we retain the Ho
- Conclusion pet owners are as happy as people who
do not own pets - This is a good example of how changing from a
one-tailed to a two-tailed test can affect your
Ho decision
15Question 3
- Dr. Bob has invented a new drug for treating
insomnia. He hopes the drug works, but it may
actually make peoples insomnia worse (it is so
new that he isnt sure which way the drug will
work). Dr. Bob asks 10 people with insomnia to
record how many hours of sleep they get without
the drug and then has them take the drug and
record their number of hours of sleep again. Dr.
Bobs data is on the next slide
16Patient Hrs. sleep before drug Hrs. sleep after drug
1 3 5
2 4 4
3 2 5
4 2 3
5 4 4
6 5 3
7 1 6
8 0 3
9 3 5
10 2 6
17Question 3
- A) State the null and alternative hypotheses
- B) State the IV, levels of IV, and DV
- C) What type of test should you use?
- D) Test Dr. Bobs hypothesis at the .01 level of
significance - E) Test Dr. Bobs hypothesis at the .05 level of
significance using the Confidence Interval
approach
18- A) State the null and alternative hypotheses
- Ho taking the drug will not change the number of
hours of sleep that insomniacs get - Ha taking the drug will make insomniacs sleep
either more or less than before they took the
drug - B) State the IV, levels of IV, and DV
- IV insomnia drug (no drug, drug)
- DV number of hours of sleep per night
- C) What type of test should you use?
- These are pairs of scores, so use Related Samples
t
19Patient Hrs. sleep before drug Hrs. sleep after drug
1 3 5
2 4 4
3 2 5
4 2 3
5 4 4
6 5 3
7 1 6
8 0 3
9 3 5
10 2 6
D
-2
0
-3
-1
0
2
-5
-3
-2
-4
SD-18
D2
4
0
9
1
0
4
25
9
4
16
SD272
20- D) Test Dr. Bobs hypothesis at the .01 level of
significance - Alpha .01 (two tail), df 9 ( of pairs 1)
- tcrit /-3.250 gt -2.714, retain Ho
- Conclusion the drug does not work it did not
change how much sleep the insomniacs get
21- E) Test Dr. Bobs hypothesis at the .05 level of
significance using the Confidence Interval
approach - Since 0 is outside the interval, we reject Ho
- Conclusion the drug affected the number of hours
of sleep that insomniacs get (it actually
increased their sleep we can tell this by
looking at the sign of D, but you simply need to
say that the drug changed hrs of sleep b/c this
is a two-tailed test)
22Question 4
- A telephone service provider has introduced a new
long distance calling program. For a flat fee of
25 per month, callers can place as many long
distance telephone calls within Canada as they
want each month. The phone company thinks that
callers will spend more time on the phone each
month if they use the new plan. They know that
average customers spend 5 hours on the phone each
month, with a standard deviation of 45 minutes. A
sample of 30 customers using the new plan spend
an average of 6 hours on the phone each month.
23Question 4
- A) State the null and alternative hypotheses
- B) What type of test should you use?
- C) Calculate the standard error
- D) Test the phone companys hypothesis at the .05
level of significance
24- A) State the null and alternative hypotheses
- Ho Customers using the new long distance phone
plan will spend the same amount of time on the
phone as regular phone customers - Ha Customers using the new long distance phone
plan will spend more time on the phone than
regular phone customers - B) What type of test should you use?
- Comparing a sample with a population, sigma is
know, so use a Z-TEST
25- C) Calculate the standard error
- D) Test the phone companys hypothesis at the .05
level of significance - Conclusion customers using the new long distance
plan spend more time on the phone than regular
customers
Note!! Sigma is given in minutes, so you need to
state it as hours (45 minutes .75 hours) the
level of measurement of the means and standard
deviation must be the same!
zcrit 1.64 lt 7.30, reject Ho
26Question 5
- Professor Z hypothesizes that there is
relationship between the degree to which people
believe in magic and their appreciation for Harry
Potter Books. He gives a random sample of 5
people a self-report measure of Belief in Magic
that produces scores ranging from 1 (Dont
believe in magic at all) to 10 (Believe very much
in magic). He also asks them to rate their
appreciation of Harry Potter books on scale from
1 (Harry Potter books are awful) to 10 (Harry
Potter books are fabulous). Is there a
relationship between belief in magic and
appreciation of Harry Potter books?
27 Belief
in Ratings of Magic scores Harry
Potter
28Correlation Arithmetic
29The Pearson r
(SC) (SU)
SCU
N
r
30The Pearson r
(21) (31)
146
5
r
31The Pearson r
651
146
5
r
32The Pearson r
146
130.2
r
33The Pearson r
15.8
r
34The Pearson r
15.8
r
35The Pearson r
15.8
r
115
192.2
207
88.2
36The Pearson r
15.8
r
14.8
26.8
37The Pearson r
15.8
r
396.64
38The Pearson r
15.8
r
19.9158
39The Pearson r
.7933 .793
r
40The Pearson r
.793
r
See Table E.2 (p.440) for n - 2 df ( 5 - 2 3
df) and an alpha (a) of .05
41The Pearson r
.793
r
The Critical r .878 r .793 Therefore, we
retain the Ho, belief in magic and appreciation
of Harry Potter books do not appear to be
significantly correlated.
42Question 6
- Chi Square Goodness of Fit Test Example (see
answer on next page) - 1. Before an Ontario provincial election,
political researchers thought that the Liberal
Party would receive 45 of the votes, the Reform
Party would receive 25, the NDP would receive
20, and the PC party would receive 10 of the
votes. After the election, they examined 500
random votes and found that the political parties
received the following number of votes - Liberal Party 200
- Reform Party 100
- NDP 150
- PC Party 50
43Question 6 continued
- 5. A) State the null and alternative hypotheses
- 5. B) Test the hypothesis at the .05 level of
significance
44Hypotheses
- Ho The votes in the election were as predicted
by the political researchers (OE) - Ha The votes in the election were distributed
differently than as predicted (O does not equal E)
45Chi Square Chart
Calculate expected frequencies (multiply each
by n 500)
Liberals Reform NDP PC
O 200 100 150 50
E 225 125 100 50
O - E -25 -25 50 0
(O E)2 625 625 2500 0
(O E)2/E 2.7778 5 25 0
Sum 32.7778 32.7778 32.7778 32.7778
46Question 6 Continued
- Chi Square observed 32.78
- DF k-1 4-1 3, alpha .05
- Chi Square critical 7.82
- B/c Chi Square obs gt Chi Square crit, we reject
the Ho. - Conclusion The political researchers were
wrong. The votes in the election were distributed
differently than as predicted.
47Question 7
- Researchers want to know if men and women get
sick at different rates when they are exposed to
the flu. They contact 100 men and 100 women who
have been in contact with someone with the flu
over the past week and record how many got sick
and how many stayed well. The results are
presented in the table below - Men
Women
55 45
45 55
Sick Stayed Well
48Question 7 Continued
- A) State the hypotheses
- B) Test the hypothesis at the .01 level of
significance
49Hypotheses
- Ho Gender and getting sick when exposed to the
flu are independent - Ha Gender and getting sick when exposed to the
flu are not independent (are dependent)
50Calculating Expected Frequencies and Chi-Square
Men Women
O 55 E 50 O 45 E 50
O 45 E 50 O 55 E 50
Sick Stayed Well
100 100
100 100 200
.5 .5 .5 .5 2.00
51Question 7 Continued
- Chi Square crit, df 1, alpha .01 6.63, so
retain Ho - Conclusion Women and men get sick from the flu
at the same rates (gender and getting sick when
exposed to the flu are independent)
52Good Luck!!