Question

The following 5 questions are based on this information: 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.
Required:
At the 5% significance level, is the economist claim supported by the data?

Answer 1

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.