An article reported that for a sample of 45 kitchens with gas cooking appliances monitored during a one-week period, the sample mean [tex]$CO_2$[/tex] level (ppm) was 654.16, and the sample standard deviation was 166.08.

(a) Calculate and interpret a [tex]$95\%[tex]$[/tex] (two-sided) confidence interval for the true average [tex]$[/tex]CO_2$[/tex] level in the population of all homes from which the sample was selected. (Round your answers to two decimal places.)

604.47 [tex]$\qquad$[/tex] 703.85 [tex]$\square$[/tex] ppm

Interpret the resulting interval:
- We are [tex]$95\%$[/tex] confident that this interval does not contain the true population mean.
- We are [tex]$95\%$[/tex] confident that the true population mean lies above this interval.
- We are [tex]$95\%$[/tex] confident that the true population mean lies below this interval.
- We are [tex]$95\%$[/tex] confident that this interval contains the true population mean.

(b) Suppose the investigators had made a rough guess of 165 for the value of [tex]$s[tex]$[/tex] before collecting data. What sample size would be necessary to obtain an interval width of [tex]$[/tex]51[tex]$[/tex] ppm for a confidence level of [tex]$[/tex]95\%$[/tex]? (Round your answer up to the nearest whole number.)

[tex]$\square$[/tex] kitchens



Answer :

Let's address each part of the problem systematically.

### Part (a): Calculate and interpret a 95% confidence interval for the true average CO2 level

Given:
- Sample size (\( n \)) = 45
- Sample mean (\( \bar{x} \)) = 654.16 ppm
- Sample standard deviation (\( s \)) = 166.08 ppm
- Confidence level = 95%

Step 1: Determine the critical value (z-value) for a 95% confidence interval

For a 95% confidence level, the corresponding z-value is approximately 1.96 (using standard z-tables or statistical functions).

Step 2: Calculate the margin of error

The formula for the margin of error (E) in the context of a confidence interval for the mean is:
[tex]\[ E = z \times \left( \frac{s}{\sqrt{n}} \right) \][/tex]

Plugging in the given values:
[tex]\[ E = 1.96 \times \left( \frac{166.08}{\sqrt{45}} \right) \][/tex]

Step 3: Calculate the confidence interval

The confidence interval is given by:
[tex]\[ \bar{x} \pm E \][/tex]

So the lower bound and upper bound are:
[tex]\[ \text{Lower bound} = 654.16 - E \approx 605.64 \][/tex]
[tex]\[ \text{Upper bound} = 654.16 + E \approx 702.68 \][/tex]

Thus, the 95% confidence interval for the true average CO2 level is approximately (605.64 ppm, 702.68 ppm).

Interpretation:

We are 95% confident that this interval contains the true population mean.

### Part (b): Determine the necessary sample size for a desired interval width

Given:
- Desired interval width = 51 ppm
- Guessed standard deviation (\( \sigma \)) = 165 ppm
- Confidence level = 95%

Step 1: Determine the critical value (z-value) for a 95% confidence interval

From part (a), the z-value for a 95% confidence level is 1.96.

Step 2: Use the margin of error formula to find the required sample size

The margin of error \( E \) is half of the desired interval width:
[tex]\[ E = \frac{51}{2} = 25.5 \][/tex]

The formula to solve for \( n \) is derived from the margin of error formula:
[tex]\[ E = z \times \left( \frac{\sigma}{\sqrt{n}} \right) \][/tex]

Rearranging to solve for \( n \):
[tex]\[ n = \left( \frac{z \times \sigma}{E} \right)^2 \][/tex]

Plugging in the given values:
[tex]\[ n = \left( \frac{1.96 \times 165}{25.5} \right)^2 \approx 160.46 \][/tex]

Since sample size must be an integer, we round up to the nearest whole number:
[tex]\[ n \approx 161 \][/tex]

Thus, the necessary sample size to achieve a desired interval width of 51 ppm at a 95% confidence level is 161 kitchens.

In conclusion:
- The 95% confidence interval for the true average CO2 level is approximately (605.64 ppm, 702.68 ppm). We are 95% confident that this interval contains the true population mean.
- To obtain a 95% confidence interval with a width of 51 ppm, the required sample size is 161 kitchens.

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