SOLUTION: One hundred people are included in a study in which they are compared to a known population that has a mean of 73, a standard deviation of 20, and a rectangular distribution. a.

Question 1182185: One hundred people are included in a study in which they are compared to a known population that has a mean of 73, a standard deviation of 20, and a rectangular distribution.
a. μM = __________.
b. σM = __________. **This answer is NOT 20!
c. The shape of the comparison distribution is __________. **This answer is NOT rectangular!
d. If the sample mean is 75, the lower limit for the 99% confidence interval is __________.
e. If the sample mean is 75, the upper limit for the 99% confidence interval is __________.
f. If the sample mean is 75, the lower limit for the 95% confidence interval is __________.
g. If the sample mean is 75, the upper limit for the 95% confidence interval is __________.

Answer by CPhill(1959) (Show Source): You can put this solution on YOUR website!
Here's how to solve this problem:
**a. μM = 73**
The mean of the sampling distribution of the mean (μM) is equal to the population mean (μ). Therefore, μM = 73.
**b. σM = 2**
The standard deviation of the sampling distribution of the mean (σM), also known as the standard error, is calculated as:
σM = σ / √n
Where:
* σ = population standard deviation = 20
* n = sample size = 100
σM = 20 / √100 = 20 / 10 = 2
**c. The shape of the comparison distribution is approximately normal.**
Even though the population distribution is rectangular, the Central Limit Theorem states that the sampling distribution of the mean will approach a normal distribution as the sample size increases. With a sample size of 100, the sampling distribution will be approximately normal.
**d. Lower limit for the 99% confidence interval:**
For a 99% confidence interval, the alpha (α) level is 1 - 0.99 = 0.01. Alpha/2 is 0.01 / 2 = 0.005. The critical z-score (zα/2) for a 99% confidence interval is approximately 2.576.
The formula for the confidence interval is:
CI = M ± zα/2 * σM
Where:
* M = sample mean = 75
Lower Limit = M - zα/2 * σM
Lower Limit = 75 - (2.576 * 2)
Lower Limit ≈ 75 - 5.152
Lower Limit ≈ 69.85
**e. Upper limit for the 99% confidence interval:**
Upper Limit = M + zα/2 * σM
Upper Limit = 75 + (2.576 * 2)
Upper Limit ≈ 75 + 5.152
Upper Limit ≈ 80.15
**f. Lower limit for the 95% confidence interval:**
For a 95% confidence interval, α = 1 - 0.95 = 0.05. Alpha/2 is 0.05 / 2 = 0.025. The critical z-score (zα/2) for a 95% confidence interval is approximately 1.96.
Lower Limit = M - zα/2 * σM
Lower Limit = 75 - (1.96 * 2)
Lower Limit ≈ 75 - 3.92
Lower Limit ≈ 71.08
**g. Upper limit for the 95% confidence interval:**
Upper Limit = M + zα/2 * σM
Upper Limit = 75 + (1.96 * 2)
Upper Limit ≈ 75 + 3.92
Upper Limit ≈ 78.92

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