To find the 90% confidence interval for the mean weight of garbage discarded by the 15 households, we will use the formula for the confidence interval of the mean when the sample standard deviation is known. The formula is:

$$ \bar{x} \pm t_{\alpha/2} \left( \frac{s}{\sqrt{n}} \right) $$

where:
- $\bar{x}$ is the sample mean
- $t_{\alpha/2}$ is the t-score for the desired confidence level
- $s$ is the sample standard deviation
- $n$ is the sample size

Given:
- $\bar{x} = 34.9$ lbs
- $s = 11.2$ lbs
- $n = 15$
- Confidence level = 90%

First, we need to find the t-score for a 90% confidence level with $n - 1 = 14$ degrees of freedom. Using a t-table or calculator, we find that $t_{0.05, 14} \approx 1.7613$.

Next, we calculate the standard error of the mean (SEM):

$$ \text{SEM} = \frac{s}{\sqrt{n}} = \frac{11.2}{\sqrt{15}} \approx 2.8919 $$

Now, we can calculate the margin of error (ME):

$$ \text{ME} = t_{\alpha/2} \times \text{SEM} = 1.7613 \times 2.8919 \approx 5.0961 $$

Finally, we calculate the confidence interval:

$$ \bar{x} - \text{ME} \leq \mu \leq \bar{x} + \text{ME} $$

$$ 34.9 - 5.0961 \leq \mu \leq 34.9 + 5.0961 $$

$$ 29.8039 \leq \mu \leq 39.9961 $$

Therefore, the 90% confidence interval for the mean weight of garbage discarded by the 15 households is approximately $29.8039$ lbs to $39.9961$ lbs.

Question

To find the 90% confidence interval for the mean weight of garbage discarded by the 15 households, we will use the formula for the confidence interval of the mean when the sample standard deviation is known. The formula is:

$$ \bar{x} \pm t_{\alpha/2} \left( \frac{s}{\sqrt{n}} \right) $$

where:
- $\bar{x}$ is the sample mean
- $t_{\alpha/2}$ is the t-score for the desired confidence level
- $s$ is the sample standard deviation
- $n$ is the sample size

Given:
- $\bar{x} = 34.9$ lbs
- $s = 11.2$ lbs
- $n = 15$
- Confidence level = 90%

First, we need to find the t-score for a 90% confidence level with $n - 1 = 14$ degrees of freedom. Using a t-table or calculator, we find that $t_{0.05, 14} \approx 1.7613$.

Next, we calculate the standard error of the mean (SEM):

$$ \text{SEM} = \frac{s}{\sqrt{n}} = \frac{11.2}{\sqrt{15}} \approx 2.8919 $$

Now, we can calculate the margin of error (ME):

$$ \text{ME} = t_{\alpha/2} \times \text{SEM} = 1.7613 \times 2.8919 \approx 5.0961 $$

Finally, we calculate the confidence interval:

$$ \bar{x} - \text{ME} \leq \mu \leq \bar{x} + \text{ME} $$

$$ 34.9 - 5.0961 \leq \mu \leq 34.9 + 5.0961 $$

$$ 29.8039 \leq \mu \leq 39.9961 $$

Therefore, the 90% confidence interval for the mean weight of garbage discarded by the 15 households is approximately $29.8039$ lbs to $39.9961$ lbs.

Knowee AI · Accepted Answer