Question 1174400: In a certain food stall, 278 out of 500 randomly selected consumers indicate their preference for a new kind of food combination. Use a 99% confidence interval to estimate the true proportion p who like the new food combination.
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! **1. State the Hypotheses:**
* **Null Hypothesis (H0):** The sample is from a population with a mean height of 171.20 cm (µ = 171.20).
* **Alternative Hypothesis (H1):** The sample is not from a population with a mean height of 171.20 cm (µ ≠ 171.20).
**2. Determine the Significance Level:**
* Alpha (α) = 0.05
**3. Calculate the Test Statistic:**
* Since we have a large sample size (n = 400) and know the population standard deviation, we'll use a z-test.
* z = (sample mean - population mean) / (population standard deviation / √sample size)
* z = (171.40 - 171.20) / (3.3 / √400) = 0.20 / 0.165 ≈ 1.21
**4. Determine the Critical Value:**
* For a two-tailed test at α = 0.05, the critical z-value is ±1.96.
**5. Compare the Test Statistic to the Critical Value:**
* Our calculated z-value (1.21) falls within the range of -1.96 to +1.96.
**6. Make a Decision:**
* Since the test statistic falls within the acceptance region, we **fail to reject the null hypothesis**.
**7. Conclusion:**
* At a 5% level of significance, there is not enough evidence to conclude that the sample of 400 males with a mean height of 171.40 cm is not from a population with a mean height of 171.20 cm and a standard deviation of 3.3. Therefore, **it can be reasonably regarded as a sample from that population.**
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