3. The mean family income in Arizona is about $64,750 with a standard deviation (for the population) of $59,750. Imagine that you are taking a sample of 200 randomly selected state residents. (a) What is the probability that your sample mean is between $61,000 and $64,750? (b) For this same sample size, what is the probability that the sample mean exceeds $75,000? (c) What income levels would correspond with the 95% confidence limits for this sample? (d) Interpret what these 95% confidence limits tell us
Since you are dealing with a distribution of means rather than raw scores, use the following equation and the same table.
Z = (score-mean)/SEm
SEm = SD/√n
95% = mean ± 1.96 SEm
31.59%
To answer these questions, we'll be using the concepts of probability, the normal distribution, and confidence intervals. Here's how we can approach each part:
(a) To find the probability that the sample mean is between $61,000 and $64,750, we need to standardize the sample mean using the formula for z-score:
z = (x - μ) / (σ / √n)
Where:
x = sample mean
μ = population mean
σ = population standard deviation
n = sample size
In this case:
x = $61,000
μ = $64,750
σ = $59,750
n = 200
First, let's find the z-scores for both $61,000 and $64,750:
z1 = ($61,000 - $64,750) / ($59,750 / √200)
z1 ≈ -0.76
z2 = ($64,750 - $64,750) / ($59,750 / √200)
z2 ≈ 0
Next, we can find the probabilities associated with these z-scores using a standard normal distribution table or a calculator. The probability we need is P(z1 < z < z2), which represents the area under the curve between the two z-scores.
P(-0.76 < z < 0) ≈ 0.285
So, the probability that the sample mean is between $61,000 and $64,750 is approximately 0.285 or 28.5%.
(b) To find the probability that the sample mean exceeds $75,000, we need to find the z-score for $75,000 and then calculate the probability of getting a z-score greater than that.
z = ($75,000 - $64,750) / ($59,750 / √200)
z ≈ 1.38
Now, we can find the probability associated with this z-score:
P(z > 1.38) ≈ 0.084
So, the probability that the sample mean exceeds $75,000 is approximately 0.084 or 8.4%.
(c) To determine the income levels corresponding to the 95% confidence limits, we'll calculate the margin of error and then construct the confidence interval.
The margin of error can be found using the following formula:
ME = z * (σ / √n)
Where:
ME = Margin of Error
z = z-score corresponding to the desired confidence level (95%)
σ = population standard deviation
n = sample size
In this case:
z = 1.96 (corresponding to a 95% confidence level)
σ = $59,750
n = 200
ME = 1.96 * ($59,750 / √200)
ME ≈ $8,376.49
The margin of error represents the maximum range of likely values around the sample mean. To construct the confidence interval, we subtract the margin of error from and add it to the sample mean:
Lower Limit = $64,750 - $8,376.49
Upper Limit = $64,750 + $8,376.49
Lower Limit ≈ $56,373.51
Upper Limit ≈ $73,126.49
Therefore, the income levels corresponding to the 95% confidence limits for this sample are approximately $56,373.51 and $73,126.49.
(d) The 95% confidence limits tell us that, based on the sample taken, we can be 95% confident that the true population mean falls within the range of approximately $56,373.51 and $73,126.49. In other words, if we were to repeat the sampling process multiple times and construct the confidence intervals each time, approximately 95% of those intervals would contain the true population mean. It provides a measure of the precision of our estimate for the population mean.