SOLUTION: d. According to an officer in the Health Care Financing Administration, the average annual expenditure on toiletries is RM110 per person in Malaysia. The data of 30 randomly se

Question 1199956: d. According to an officer in the Health Care Financing Administration, the average annual
expenditure on toiletries is RM110 per person in Malaysia. The data of 30 randomly
selected persons are as follows and the standard deviation of the sample is 19.2152.
125 110 110 90 90
85 125 110 85 120
130 125 130 90 130
80 120 90 110 90
90 85 100 130 80
90 100 125 150 100
i. Find the sample mean of the data.
ii. Is there enough evidence to support the officer’s claim at 0.02 level of significance?
iii. What is the decision at a 0.2% significance level?
Answer by textot(100) (Show Source): You can put this solution on YOUR website!
**i. Find the sample mean of the data.**
1. **Sum of all expenditures:**
125 + 110 + 110 + 90 + 90 + 85 + 125 + 110 + 85 + 120 + 130 + 125 + 130 + 90 + 130 + 80 + 120 + 90 + 110 + 90 + 90 + 85 + 100 + 130 + 80 + 90 + 100 + 125 + 150 + 100 = 3331
2. **Calculate the sample mean (x̄):**
x̄ = Sum of all expenditures / Number of observations
x̄ = 3331 / 30
x̄ = 111.03
**Therefore, the sample mean expenditure on toiletries is RM111.03.**
**ii. Is there enough evidence to support the officer’s claim at 0.02 level of significance?**
1. **Set up hypotheses:**
- **Null Hypothesis (H0):** μ = 110 (The average annual expenditure is RM110)
- **Alternative Hypothesis (H1):** μ ≠ 110 (The average annual expenditure is not RM110)
2. **Calculate the test statistic (t-score):**
t = (x̄ - μ) / (s / √n)
t = (111.03 - 110) / (19.2152 / √30)
t ≈ 0.29
3. **Determine degrees of freedom:**
Degrees of freedom (df) = n - 1 = 30 - 1 = 29
4. **Find the critical values:**
- Since this is a two-tailed test with α = 0.02, we need to find the critical t-values that correspond to the 0.01 level of significance in each tail (0.02 / 2 = 0.01).
- Using a t-distribution table or a calculator, find the critical t-values for df = 29 and α/2 = 0.01.
- The critical values will be approximately ±2.756.
5. **Decision:**
- Since the calculated t-value (0.29) falls within the non-rejection region (-2.756 < t < 2.756), we **fail to reject the null hypothesis**.
**Conclusion:**
At the 0.02 level of significance, there is **not enough evidence** to reject the officer's claim that the average annual expenditure on toiletries is RM110 per person in Malaysia.
**iii. What is the decision at a 0.2% significance level?**
* 0.2% significance level is equivalent to α = 0.002.
* For α = 0.002 (two-tailed), the critical t-values will be more extreme (further from 0) compared to α = 0.02.
* Since the calculated t-value (0.29) is still within the non-rejection region for α = 0.002, the decision remains the same: **fail to reject the null hypothesis**.
**In summary:**
* The sample mean expenditure is RM111.03.
* At both 0.02 and 0.002 significance levels, there is not enough evidence to reject the officer's claim that the average annual expenditure on toiletries is RM110 per person in Malaysia.

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