a) The parameter we are estimating is the true average amount spent on a Halloween costume for U.S. adults aged 18-24.

b) We can work this problem because the sample size is sufficiently large (n = 41), which allows us to use the Central Limit Theorem to assume that the sampling distribution of the sample mean is approximately normal.

c) The point estimate of that parameter is the sample mean, which is $43.3.

d) The standard deviation of the sampling distribution for the sample means (also known as the standard error) is calculated using the formula:
$$ \text{Standard Error} = \frac{\text{Standard Deviation}}{\sqrt{n}} $$
$$ \text{Standard Error} = \frac{14.9}{\sqrt{41}} \approx 2.3261 $$

e) To find the margin of error for a 90% confidence interval, we need the critical value (z*) for a 90% confidence level. For a 90% confidence interval, the critical value (z*) is approximately 1.645. The margin of error (ME) is calculated as:
$$ \text{ME} = z^* \times \text{Standard Error} $$
$$ \text{ME} = 1.645 \times 2.3261 \approx 3.8268 $$

f) We are 90% confident that the true average amount spent on a Halloween costume for U.S. adults aged 18-24 is between:
$$ \text{Lower Limit} = \text{Sample Mean} - \text{ME} $$
$$ \text{Lower Limit} = 43.3 - 3.8268 \approx 39.4732 $$

$$ \text{Upper Limit} = \text{Sample Mean} + \text{ME} $$
$$ \text{Upper Limit} = 43.3 + 3.8268 \approx 47.1268 $$

Therefore, we are 90% confident that the true average amount spent on a Halloween costume for U.S. adults aged 18-24 is between $39.4732 and $47.1268.

Question

a) The parameter we are estimating is the true average amount spent on a Halloween costume for U.S. adults aged 18-24.

b) We can work this problem because the sample size is sufficiently large (n = 41), which allows us to use the Central Limit Theorem to assume that the sampling distribution of the sample mean is approximately normal.

c) The point estimate of that parameter is the sample mean, which is $43.3.

d) The standard deviation of the sampling distribution for the sample means (also known as the standard error) is calculated using the formula:
$$ \text{Standard Error} = \frac{\text{Standard Deviation}}{\sqrt{n}} $$
$$ \text{Standard Error} = \frac{14.9}{\sqrt{41}} \approx 2.3261 $$

e) To find the margin of error for a 90% confidence interval, we need the critical value (z*) for a 90% confidence level. For a 90% confidence interval, the critical value (z*) is approximately 1.645. The margin of error (ME) is calculated as:
$$ \text{ME} = z^* \times \text{Standard Error} $$
$$ \text{ME} = 1.645 \times 2.3261 \approx 3.8268 $$

f) We are 90% confident that the true average amount spent on a Halloween costume for U.S. adults aged 18-24 is between:
$$ \text{Lower Limit} = \text{Sample Mean} - \text{ME} $$
$$ \text{Lower Limit} = 43.3 - 3.8268 \approx 39.4732 $$

$$ \text{Upper Limit} = \text{Sample Mean} + \text{ME} $$
$$ \text{Upper Limit} = 43.3 + 3.8268 \approx 47.1268 $$

Therefore, we are 90% confident that the true average amount spent on a Halloween costume for U.S. adults aged 18-24 is between $39.4732 and $47.1268.

Knowee AI · Accepted Answer