Question 1171064
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.