Question:The question is incomplete. Find below the options and the answers.
1. To test the economist’s claim, the competing hypotheses should be formulated as
Select one:
a. H0:μ1-μ2>0 versus Ha:μ1-μ2≤0
b. H0:μ1-μ2≥0 versus Ha:μ1-μ2<0
c. H0:μ1-μ2≤0 versus Ha:μ1-μ2>0
2.
The standard error of x(1)bar- x(2) bar is
Select one:
a. 0.82
b. 2.70
c. 12.5
d. 9.25
3. The value of the test statistics is
Select one:
a. 0.40
b. 1.85
c. 0.54
d. 27.78
4. The p-value of the test is
Select one:
a. 0.34
b. 0.03
c. 0.29
d. 0.08
5. At α=0.05,
Select one:
a. We can reject H(0) in favor of H(a)
b. We cannot reject H(0)
c. We can conclude that average weekly food expenditures in City 1 is less than that of City 2
Answer:
1. The correct answer is option (a) H0:μ1-μ2>0 versus Ha:μ1-μ2≤0
2. The correct answer is option b. 2.70
3. The correct answer is option b. 1.85
4. The correct answer is option b. 0.03
5. The correct answer is option c. We can conclude that average weekly food expenditures in City 1 is less than that of City 2
Step-by-step explanation:
Given Data:
City 1 City 2
x1(bar)=164 x2 (bar) =159
σ(1)=12.5 σ(2) =9.25
n(1)=35 n2=30
2. The standard error of x(1)bar- x(2) bar is calculated using the formula;
Standard error = √σ₁² +σ₂² /n₁+n₂
= √(12.5²+9.25²/35 +30)
= 2.7
3. The test statistics is calculated using the formula;
z = x(1)bar- x(2)/Standard error
= 164 - 159/2.7
= 5/2.7
= 1.85
4. using the normal standard table, the p- value at z = 1.85 is 0.03
5. We can conclude that average weekly food expenditures in City 1 is less than that of City 2