Question 1183928: The data below give the lifetimes in hundreds of hours of samples of two types of electronic tubes. Past lifetime data of such tubes have shown that they can often be modeled as arising from a lognormal distribution. That is, the logarithms of the data are normally distributed. Assuming that variance of the logarithms is equal for the two populations, test, at the 5 percent level of significance, the hypothesis that the two population distributions are identical. Type 1 32, 84, 37, 42, 78, 62, 59, 74 Type 2 39, 111, 55, 106, 90, 87, 85
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Here's how to perform the hypothesis test:
**1. Transform the Data:**
Since the lifetimes are lognormally distributed, we first take the natural logarithm (ln) of each data point. This will give us data that is approximately normally distributed.
**Type 1 (ln transformed):** 3.466, 4.430, 3.611, 3.738, 4.356, 4.127, 4.089, 4.304
**Type 2 (ln transformed):** 3.664, 4.710, 4.007, 4.663, 4.499, 4.465, 4.442
**2. State the Hypotheses:**
* **Null Hypothesis (H0):** The two population distributions are identical (i.e., the means of the log-transformed data are equal). μ₁ = μ₂
* **Alternative Hypothesis (H1):** The two population distributions are not identical (i.e., the means of the log-transformed data are not equal). μ₁ ≠ μ₂ (This is a two-tailed test).
**3. Significance Level:** α = 0.05
**4. Calculate Sample Statistics:**
Calculate the sample mean (x̄) and sample variance (s²) for each transformed sample.
**Type 1:**
* n₁ = 8
* x̄₁ ≈ 4.015
* s₁² ≈ 0.111
**Type 2:**
* n₂ = 7
* x̄₂ ≈ 4.343
* s₂² ≈ 0.165
**5. Perform the t-test:**
Since the population variances are assumed equal, we use the pooled variance t-test.
* **Pooled Variance (sₚ²):**
sₚ² = [(n₁ - 1)s₁² + (n₂ - 1)s₂²] / (n₁ + n₂ - 2)
sₚ² = [(7 * 0.111) + (6 * 0.165)] / (8 + 7 - 2)
sₚ² ≈ 0.136
* **Standard Error:**
SE = sqrt[sₚ²(1/n₁ + 1/n₂)]
SE = sqrt[0.136(1/8 + 1/7)]
SE ≈ 0.187
* **t-statistic:**
t = (x̄₁ - x̄₂) / SE
t = (4.015 - 4.343) / 0.187
t ≈ -1.751
* **Degrees of Freedom:**
df = n₁ + n₂ - 2 = 8 + 7 - 2 = 13
**6. Determine the Critical Value:**
For a two-tailed test at α = 0.05 and df = 13, the critical t-value is approximately ±2.160 (you'll need a t-table or calculator).
**7. Make a Decision:**
Our calculated t-statistic (-1.751) falls *within* the range of -2.160 to +2.160. Therefore, we *fail to reject* the null hypothesis.
**8. Conclusion:**
At the 5% level of significance, there is *not* sufficient evidence to conclude that the two population distributions of electronic tube lifetimes are different. While the sample means are different, the difference is not statistically significant given the variability in the data.
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