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Answer:
>> Null hypothesis; H0: μ1 - μ2 ≥ 0
Null hypothesis; Ha: μ1 - μ2 < 0
>> z = 1.85
>> p-value = 0.03
>> there is evidence sufficient to reject the claim that average weekly food expenditure of households in City 1 is more than that of households in City 2.
The economist claim is not supported by data.
Step-by-step explanation:
We are given;
Sample mean of city 1; x1¯ = 164
Sample mean of city 2; x2¯ = 159
Sample size of city 1; n1 = 35
Sample size of city 2; n2 = 30
Standard deviation of city 1; σ1 = 12.5
Standard deviation of city 2; σ2 = 9.25
Let's define the hypotheses;
Null hypothesis; H0: μ1 - μ2 ≥ 0
Null hypothesis; Ha: μ1 - μ2 < 0
Test statistic which is the z-score Formula between 2 mean is;
z = ((x1¯ - x2¯) - 0)/√((σ1)²/n1) + ((σ2)²/n2))
z = (164 - 159 - 0)/√((12.5²/35) + (9.25²/30))
z = 5/2.705
z = 1.85
From online p-value from z-score calculator attached, using z = 1.85, one tailed, significance value of 0.05, we have;
p-value = 0.03
The p-value is less than the significance value, and so we will reject the null hypothesis and conclude that there is evidence sufficient to reject the claim that average weekly food expenditure of households in City 1 is more than that of households in City 2.
