Download Introduction to the Practice of Statistics

Introduction to the Practice of Statistics
Fifth Edition
Moore, McCabe
Section 7.1 Homework Answers
7.13 For a sample of size 5, a test of a null hypothesis versus a two-sided alternative gives t= 2.45
(a) Is the test result significant at the 5% level? Draw a sketch of the appropriate t distribution and
illustrate your calculation with this
sketch.
P(t > 2.45) = 0.03522
The p-value = 2(0.03522)
= 0.070445
Used Excel “tdist(2.45, 4, 1)
The result is not significant at 5%, since my p-value is larger; 7.045% > 5%
(b) Now assume that the same statistic was obtained for a sample size of n =10. Assess the statistical
significance of the result and illustrate the calculation with a sketch. How did the statistical
significance change with the sample size? Explain your answer.
P(t > 2.45) = 0.01838
The p-value = 2(0.01838)
= 0.03676
Used Excel “tdist(2.45, 9, 1)
The result is significant at 5%.
7.15 Repeat the previous exercise for the two situations where the alternative is
one-sided.
n = 2000. α = 1%, since we are looking at either side, I will draw two graphs.
I used Excel to get the desired result; tinv(0.02, 1999).
−4
−3
−2
−1
0
1
2
3
4
t5
2.328
7.16 Computer software reports x = 12.3 and P = 0.08 for a t test of Ho: µ = 0 versus Ha: µ ≠ 0.
Base on prior knowledge, you can justify testing the alternative Ha: µ > 0. What is the P-value for
your significance test?
The original p-value is based on doubling the following calculation P( x > 12.3)
p-value is 2(P( x > 12.3) ) = 0.08.
A one-sided test is just half the value. Thus, the p-value is 0.04.
Two-sided test
x = 12.3
One-sided test
x = 12.3
7.17
Suppose that x = -12.3 in the setting of the previous exercise. Would this change your
answer? Use a sketch of the distribution of the test statistic under the null hypothesis to
illustrate and explain your answer.
The answer all depends on the alternative hypothesis.
If we know change the alternative to
Ha µ < 0, then (P( x <- 12.3) = 0.04.
x = -12.3
µ= 0
X
Otherwise, if the alternative continues to be
Ha: µ > 0, then P( x > -12.3) = 0.96, the complement.
x = -12.3
µ= 0
X
7.21 The hypotheses Ho: µ = 50 and Ha: µ ≠ 50 are examined using a sample of size n = 30. The
one-sample t statistic has the value t = 1.35.
(a) Give the degrees of freedom for the test statistic.
The degrees of freedom is 29.
(b) Locate the two critical values t* from Table D that bracket t. What are the right-tail probabilities
p for these two values?
When I look at the table in the back, I will only look at the row labeled 29 for the correct degrees of
freedom.
So my t-value of 1.350 is between 1.311 and 1.699; 1.311 < 1.350 < 1.699. Thus if I was
calculating the probability of P(t > 1.35) all the information the table provides is that
0.05 < P(t > 1.35) < 0.1
If I was calculating a p-value then I know that I need to double the stated probabilities my p-value is
between 10% and 20%.
2(0.05) < 2(P(t > 1.35)) < 2( 0.1)
0.1 < p-value < 0.2
(c) How would you report the P-value for this test?
The hypotheses Ho: µ = 50 and Ha: µ ≠ 50 (two-sided test)
If I was calculating a p-value then I know that I need to double the stated probabilities my p-value is
between 10% and 20%.
2(0.05) < 2(P(t > 1.35)) < 2( 0.1)
0.1 < p-value < 0.2
(d) Is the value t = 1.35 statistically significant at the 10% level? At the 5%
level?
The result is not statistically significant for either the 5% nor the 10% level, for the two-sided test of
significance.
(e) Illustrate your answers to the previous parts of this exercise with a sketch of the t distribution.
α = 10%
α = 5%
(f) If you have software available, find the exact P-value.
Using Excel (tdist command), I found the p-value, =tdist(1.35, 29, 2), to be 0.1875.