Math Problem Statement

An article reported that for a sample of 48 kitchens with gas cooking appliances monitored during a one-week period, the sample mean CO2 level (ppm) was 654.16, and the sample standard deviation was 165.4.

(a) Calculate and interpret a 95% (two-sided) confidence interval for true average CO2 level in the population of all homes from which the sample was selected. (Round your answers to two decimal places.)

  ,  

 ppm

Interpret the resulting interval.

We are 95% confident that this interval contains the true population mean.We are 95% confident that this interval does not contain the true population mean.    We are 95% confident that the true population mean lies above this interval.We are 95% confident that the true population mean lies below this interval.

(b) Suppose the investigators had made a rough guess of 165 for the value of s before collecting data. What sample size would be necessary to obtain an interval width of 48 ppm for a confidence level of 95%? (Round your answer up to the nearest whole number.)  kitchens

Solution

(a) Calculating the 95% Confidence Interval for the True Mean CO₂ Level

To calculate a two-sided confidence interval for the population mean, we use the formula:

CI=xˉ±tα/2sn\text{CI} = \bar{x} \pm t_{\alpha/2} \cdot \frac{s}{\sqrt{n}}

Where:

  • xˉ=654.16\bar{x} = 654.16 (sample mean)
  • s=165.4s = 165.4 (sample standard deviation)
  • n=48n = 48 (sample size)
  • tα/2t_{\alpha/2}: critical value for 95% confidence level with df=n1=47df = n - 1 = 47.

Step 1: Find tα/2t_{\alpha/2}

Using a tt-distribution table or software, the tα/2t_{\alpha/2} value for df=47df = 47 at 95% confidence is approximately 2.0112.011.

Step 2: Calculate the Margin of Error (ME)

ME=tα/2sn\text{ME} = t_{\alpha/2} \cdot \frac{s}{\sqrt{n}} ME=2.011165.448\text{ME} = 2.011 \cdot \frac{165.4}{\sqrt{48}} ME=2.01123.88148.03\text{ME} = 2.011 \cdot 23.881 \approx 48.03

Step 3: Confidence Interval

CI=654.16±48.03\text{CI} = 654.16 \pm 48.03 CI=(606.13,702.19)\text{CI} = (606.13, 702.19)

Interpretation: We are 95% confident that the true average CO₂ level in the population of homes is between 606.13 ppm and 702.19 ppm.


(b) Determining Sample Size for Desired Interval Width

The formula for the margin of error in a confidence interval is: ME=tα/2sn\text{ME} = t_{\alpha/2} \cdot \frac{s}{\sqrt{n}} To achieve a desired interval width W=48W = 48, the margin of error is half the width: ME=W2=482=24\text{ME} = \frac{W}{2} = \frac{48}{2} = 24

Step 1: Rearrange the Formula for nn

n=(tα/2sME)2n = \left(\frac{t_{\alpha/2} \cdot s}{\text{ME}}\right)^2

Step 2: Plug in Values

Using tα/2=2.011t_{\alpha/2} = 2.011, s=165s = 165, and ME=24\text{ME} = 24: n=(2.01116524)2n = \left(\frac{2.011 \cdot 165}{24}\right)^2 n=(331.81524)2(13.826)2191.23n = \left(\frac{331.815}{24}\right)^2 \approx (13.826)^2 \approx 191.23

Since sample size must be an integer, round up to the nearest whole number: n=192n = 192

Answer: A sample size of 192 kitchens would be necessary to achieve a 95% confidence interval width of 48 ppm.


Recap and Questions

  • (a) 95% Confidence Interval: 606.13ppmto702.19ppm606.13 \, \text{ppm} \, \text{to} \, 702.19 \, \text{ppm}
  • (b) Required Sample Size: n=192n = 192
  1. Why is the tt-distribution used instead of the normal distribution here?
  2. What factors influence the width of a confidence interval?
  3. How does increasing the sample size affect the confidence interval?
  4. What assumptions are made when calculating the confidence interval?
  5. How would the interval change if the sample standard deviation were smaller?

Tip: Confidence intervals provide a range of plausible values for the population mean, but they rely on the assumption of a representative sample.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Statistics
Confidence Interval
Sample Size Determination
t-Distribution

Formulas

CI = x̄ ± t_{α/2} * (s / √n)
Margin of Error (ME) = t_{α/2} * (s / √n)
Sample Size Formula: n = (t_{α/2} * s / ME)^2

Theorems

Central Limit Theorem
Properties of the t-Distribution

Suitable Grade Level

Undergraduate Statistics