An economist claims that average weekly food expenditure of households in city 1 is more than that of households in city 2. she surveys 35 households in city 1 and obtains an average weekly food expenditure of $164. a sample of 30 households in city 2 yields an average weekly expenditure of $159. historical data reveals that the population standard deviation for city 1 and city 2 are $12.50 and $9.25, respectively. to test the economist's claim, the competing hypotheses should be formulated as

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

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

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

the third answer is correct.