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Math  /  Data & Statistics

QuestionTo encourage water conservation, a newspaper is publishing an article about xeriscaping, a method of landscaping that requires less water. For the article, a newspaper reporter contacted 325 randomly chosen homeowners, and asked them what kind of lawns they have. \begin{tabular}{|l|c|} \hline Lawn type & Homeowners \\ \hline natural grass & 43 \\ \hline artificial turf & 275 \\ \hline rock & 2 \\ \hline no lawn area & 5 \\ \hline \end{tabular}
Find a 95%95 \% confidence interval for pp, the proportion of homeowners with natural grass lawns. Round your answers to the nearest thousandth. □\square <p<<p< □\square

Studdy Solution

STEP 1

What is this asking? Out of 325 homeowners, 43 have natural grass lawns.
We need to find a range where we're 95% sure the *true* proportion of natural grass lawns falls. Watch out! Don't mix up the number of homeowners with natural grass lawns (43) with the *total* number of homeowners surveyed (325)!

STEP 2

1. Calculate the sample proportion.
2. Find the margin of error.
3. Calculate the confidence interval.

STEP 3

Let's **dive in**!
We're given that 43 out of 325 homeowners have natural grass lawns.
We need to figure out what proportion this is.
This is our **sample proportion**, often denoted as p^\hat{p}.

STEP 4

To **calculate** p^\hat{p}, we simply **divide** the number of natural grass lawns by the total number of homeowners surveyed: p^=Number of natural grass lawnsTotal number of homeowners=43325≈0.132 \hat{p} = \frac{\text{Number of natural grass lawns}}{\text{Total number of homeowners}} = \frac{43}{325} \approx \textbf{0.132} So, our **sample proportion** is approximately 0.132\textbf{0.132}.
This means about 13.2% of the homeowners in our sample have natural grass lawns.

STEP 5

Now for the **margin of error**!
This tells us how much our sample proportion might differ from the *true* proportion.
For a 95% confidence interval, we use a **z-score** of approximately 1.96\textbf{1.96}.
This z-score corresponds to the 95% confidence level, meaning that 95% of the area under the normal distribution curve lies within 1.96 standard deviations of the mean.

STEP 6

The **formula** for the margin of error (ME) is: ME=z⋅p^(1−p^)n ME = z \cdot \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} Where: * zz is the **z-score** (1.96\textbf{1.96} for 95% confidence) * p^\hat{p} is our **sample proportion** (0.132\textbf{0.132}) * nn is the **sample size** (325\textbf{325})

STEP 7

Let's **plug in** our values: ME=1.96⋅0.132(1−0.132)325 ME = 1.96 \cdot \sqrt{\frac{0.132(1 - 0.132)}{325}} ME=1.96⋅0.132⋅0.868325 ME = 1.96 \cdot \sqrt{\frac{0.132 \cdot 0.868}{325}} ME=1.96⋅0.114576325 ME = 1.96 \cdot \sqrt{\frac{0.114576}{325}} ME≈1.96⋅0.00035254 ME \approx 1.96 \cdot \sqrt{0.00035254} ME≈1.96⋅0.018776 ME \approx 1.96 \cdot 0.018776 ME≈0.037 ME \approx \textbf{0.037} So, our **margin of error** is approximately 0.037\textbf{0.037}.

STEP 8

Almost there!
To find the **confidence interval**, we simply **subtract** and **add** the margin of error to our sample proportion.

STEP 9

**Lower bound:** p^−ME=0.132−0.037=0.095 \hat{p} - ME = 0.132 - 0.037 = \textbf{0.095}

STEP 10

**Upper bound:** p^+ME=0.132+0.037=0.169 \hat{p} + ME = 0.132 + 0.037 = \textbf{0.169}

STEP 11

Therefore, the 95% confidence interval for the proportion of homeowners with natural grass lawns is approximately 0.095<p<0.169\textbf{0.095} < p < \textbf{0.169}.

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