Question 1171064: The average daily expenditure on the food of a middle-income family is P220, with a standard deviation of P15. A sample of 10 families showed the following expenditure: P220, P190, P205, P180, P215, P200, P187, P195, P182, and P201. Test the hypothesis that the average daily expenditure is still 200. Use 0.01 (one-tailed) level.
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! Let's break down this hypothesis test step-by-step.
**1. State the Hypotheses**
* We are testing if the average daily expenditure is still 220, so we will use 220 as our population mean.
* Since the problem does not specify a direction, we will test if the mean is different from 220. This indicates a two-tailed test.
* **Null Hypothesis (H₀):** The average daily expenditure is P220.
* H₀: μ = 220
* **Alternative Hypothesis (H₁):** The average daily expenditure is not P220.
* H₁: μ ≠ 220 (two-tailed test)
**2. Calculate the Sample Mean and Standard Deviation**
* Sample data: 220, 190, 205, 180, 215, 200, 187, 195, 182, 201
* Sample size (n) = 10
* Calculate the sample mean (x̄):
* x̄ = (220 + 190 + 205 + 180 + 215 + 200 + 187 + 195 + 182 + 201) / 10 = 1975 / 10 = 197.5
* Calculate the sample standard deviation (s):
* First, calculate the squared differences from the mean:
* (220 - 197.5)^2 = 506.25
* (190 - 197.5)^2 = 56.25
* (205 - 197.5)^2 = 56.25
* (180 - 197.5)^2 = 306.25
* (215 - 197.5)^2 = 306.25
* (200 - 197.5)^2 = 6.25
* (187 - 197.5)^2 = 110.25
* (195 - 197.5)^2 = 6.25
* (182 - 197.5)^2 = 240.25
* (201 - 197.5)^2 = 12.25
* Sum of squared differences = 1606.5
* Sample variance (s^2) = 1606.5 / (10 - 1) = 1606.5 / 9 = 178.5
* Sample standard deviation (s) = √178.5 ≈ 13.36
**3. Determine the Test Statistic**
* Since the population standard deviation is unknown and the sample size is small (n < 30), we will use a t-test.
* t = (x̄ - μ) / (s / √n)
* t = (197.5 - 220) / (13.36 / √10)
* t = -22.5 / (13.36 / 3.162)
* t = -22.5 / 4.225
* t ≈ -5.326
**4. Determine the Critical Value**
* Significance level (α) = 0.01
* Degrees of freedom (df) = n - 1 = 10 - 1 = 9
* Type of test: Two-tailed
* Using a t-table or calculator, we find the critical t-values for α = 0.01 and df = 9.
* Critical t-values ≈ ±3.250
**5. Make a Decision**
* Calculated t-statistic: -5.326
* Critical t-values: ±3.250
* Since |-5.326| > 3.250, we reject the null hypothesis.
**6. Conclusion**
There is sufficient evidence at the 0.01 significance level to conclude that the average daily expenditure on food is different from P220.
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