Since we are performing a left
tailed test, the corresponding
p

value is 0.0071.
As
p
value < 0.01, we reject H
0
and accept H
1
.
Therefore, we conclude that
there was an improvement in the
cost cutting measures.
Step 5: Make a decision and interpret the result.
Excel Output
:
tTest for Mean
Cost
Mean
57.2866667
Variance
14.0640952
Observations
15
Pearson Correlation
#DIV/0!
Hypothesized Mean Difference
60
df
14
t Stat
2.8021597
P(T<=t) onetail
0.00706009
t Critical onetail
2.62449407
P(T<=t) twotail
0.01412017
t Critical twotail
2.97684273
Activity 3.5
•
A random sample of 400 families who planned to buy a
vacation residence revealed that 228 families want to
buy a condominium in Florida. At the 0.01 significance
level, test the statement that 55% of those families who
plan to purchase a vacation residence want a
condominium in Florida. To decide if the sample data
support the 55%, state your decision in terms of the null
hypothesis. Use a 0.01 level of significance.
Activity 3.5  Answer
Step 1: State the null hypothesis and the alternate hypothesis.
Claim: 55% of those families who plan to purchase a vacation residence want a
condominium in Florida (i.e.
𝜋
=0.55)
H
0
:
𝜋 = 0.55
H
1
:
𝜋 ≠ 0.55
Step 2: Select the level of significance.
α
= 0.01 as stated in the problem
Step 3: Select the test statistic.
Use
z
distribution for the proportion
Activity 3.5  Answer
Step 4: Formulate the decision rule.
Reject the null if the Zstatistic is outside the critical region.
The Zstatistic is
The critical zstatistic for a two
tailed hypothesis test with α = 0.01 is ±
2.576
Step 5: Make a decision and interpret the result.
Since the test statistic (+0.804) is not greater than +2.576, we fail to reject the
null hypothesis.
BUS105 Statistics
Seminar 4
Topics
Unit 2
•
1 Point Estimation and Confidence Intervals
•
2.1 Hypothesis Tests: OneSample
•
2.2 Hypothesis Tests: TwoSample
•
3 Analysis of Variance (ANOVA)
Review: Hypothesis Tests: TwoSample
•
Twosample tests of hypothesis
–
About mean difference
•
Independent samples
–
Known/Unknown standard deviation
•
Paired
–
About proportion difference
Recall: Hypothesis Testing Steps
Activity 4.1
–
Paired
t
Test
Nickel Savings and Loan
wishes to compare the
two companies it uses to
appraise the value of
residential homes. Nickel
Savings selected a
sample of 10 residential
properties and scheduled
both firms for an
appraisal. The results,
reported in $000, are
shown on the table (right).
At the 0.05 significance level, can we conclude there is a
difference in the mean appraised values of the homes?
Activity 4.1 Cont’d
Tasks:
In groups of 4 persons,
1.Open the Activity4.1_Data.xls file
2.Follow the 5steps procedure to conduct an appropriate hypothesis testing.
3.
Discuss whether there was a difference in the mean appraised values of the
homes.
You may refer to “Excel Self Practice Prior Seminar 3” Practice 2.4..
Activity 4.1 Answer
TwoSample Paired tTest for Mean
Step 1:
State the null and alternate hypotheses.