Question

out of a sample of 150 people who walked into a fast food restaurant, 135 said they prefer ketchup to mustard on their hamburgers. with 89% confidence, what is the confidence interval for the population mean of customers that prefer ketchup to mustard on their hamburgers?\\(\bigcirc\\) ci = (83.69%, 96.31)\\(\bigcirc\\) ci = (85.20%, 94.80%)\\(\bigcirc\\) ci = (86.09%, 93.91%)\\(\bigcirc\\) ci = (82.61%, 95.87%)

Explanation:

Step1: Identify the sample proportion and sample size

The sample size \( n = 150 \), and the number of people who prefer ketchup is \( x = 135 \). So the sample proportion \( \hat{p}=\frac{x}{n}=\frac{135}{150} = 0.89 \) (or 89%).

Step2: Determine the confidence level (assuming 95% confidence, common if not specified, but let's check the options. Wait, the problem says "With 89% confidence"? Wait, no, maybe a typo? Wait, the options are confidence intervals. Wait, maybe it's a 95% confidence interval? Wait, let's recall the formula for confidence interval for a proportion: \( \hat{p}\pm z_{\alpha/2}\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} \)

First, calculate the standard error \( SE=\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}=\sqrt{\frac{0.89\times(1 - 0.89)}{150}}=\sqrt{\frac{0.89\times0.11}{150}}=\sqrt{\frac{0.0979}{150}}\approx\sqrt{0.0006527}\approx0.0255 \)

For a 95% confidence interval, \( z_{\alpha/2}=1.96 \)

Margin of error \( ME = z_{\alpha/2}\times SE=1.96\times0.0255\approx0.04998\approx0.05 \)

Lower bound: \( 0.89 - 0.05 = 0.84 \)? Wait, no, wait the options are (86.09%, 93.91%), (85.20%, 94.80%), (83.69%, 96.31%), (82.61%, 95.87%). Wait, maybe I made a mistake. Wait, maybe the sample proportion is 89%, and we need to find the confidence interval. Wait, let's check the option \( CI=(86.09\%, 93.91\%) \). Let's calculate the midpoint: \( \frac{86.09 + 93.91}{2}=90 \)? No, 89? Wait, 86.09 + 93.91 = 180, 180/2=90. Wait, maybe the confidence level is 90%? Wait, \( z_{\alpha/2} \) for 90% is 1.645.

\( SE=\sqrt{\frac{0.89\times0.11}{150}}\approx0.0255 \)

\( ME = 1.645\times0.0255\approx0.0419 \)

Lower bound: \( 0.89 - 0.0419 = 0.8481\approx84.81\% \)

Upper bound: \( 0.89 + 0.0419 = 0.9319\approx93.19\% \)

Wait, the option \( CI=(86.09\%, 93.91\%) \) – wait, maybe my calculation is wrong. Wait, let's recalculate:

\( \hat{p}=0.89 \), \( n = 150 \)

\( SE=\sqrt{\frac{0.89\times0.11}{150}}=\sqrt{\frac{0.0979}{150}}\approx\sqrt{0.0006526667}\approx0.02555 \)

For 95% CI, \( z = 1.96 \), \( ME = 1.96\times0.02555\approx0.0499 \)

Lower: \( 0.89 - 0.0499 = 0.8401\approx84.01\% \)

Upper: \( 0.89 + 0.0499 = 0.9399\approx93.99\% \), which is close to \( (86.09\%, 93.91\%) \)? No, 86.09 and 93.91: midpoint is 90? Wait, maybe the sample size is different? Wait, the problem says "Out of a sample of 150 people who walked into a fast food restaurant, 135 said they prefer ketchup". So \( \hat{p}=135/150 = 0.89 \). Wait, maybe the confidence interval is calculated as follows:

Wait, let's check the option \( CI=(86.09\%, 93.91\%) \). Let's compute the margin of error for this interval: \( (93.91 - 86.09)/2 = 3.91 \). So \( ME = 3.91\% = 0.0391 \)

\( ME = z_{\alpha/2}\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} \)

\( 0.0391 = z_{\alpha/2}\times\sqrt{\frac{0.89\times0.11}{150}} \)

\( \sqrt{\frac{0.89\times0.11}{150}}\approx0.0255 \)

\( z_{\alpha/2}=0.0391 / 0.0255\approx1.53 \), which is close to 94% confidence? Wait, maybe the correct option is \( CI=(86.09\%, 93.91\%) \) because when we calculate with \( \hat{p}=0.89 \), and maybe a different z-score. Wait, let's check the midpoint of each option:

- Option 1: (83.69, 96.31): midpoint 90
- Option 2: (85.20, 94.80): midpoint 90
- Option 3: (86.09, 93.91): midpoint 90
- Option 4: (82.61, 95.87): midpoint 89.24

Wait, the sample proportion is 89%, so the midpoint should be close to 89%. Wait, 86.09 and 93.91: midpoint is (86.09 + 93.91)/2 = 90. 85.20 + 94.80 = 180, midpoint 90. 83.69 + 96.31 = 180, midpoint 90. 82.61 + 95.87 = 178.48, midpoint ~89.24. Wait, maybe the question has a typo, and the sample proportion is 90%? No, 135/150 is 90%? Wait, 135 divided by 150: 135/150 = 0.9, which is 90%! Oh! I made a mistake earlier. 135/150 = 0.9, not 0.89. Oh! That's the error. 135 divided by 150: 135 ÷ 150 = 0.9, so \( \hat{p}=0.9 \) (90%).

Now, recalculate:

\( SE=\sqrt{\frac{0.9\times0.1}{150}}=\sqrt{\frac{0.09}{150}}=\sqrt{0.0006}=0.0245 \)

For 95% confidence, \( z_{\alpha/2}=1.96 \)

\( ME = 1.96\times0.0245\approx0.04802 \)

Lower bound: \( 0.9 - 0.04802 = 0.85198\approx85.20\% \)

Upper bound: \( 0.9 + 0.04802 = 0.94802\approx94.80\% \)

Ah! So that's the option \( CI=(85.20\%, 94.80\%) \). Wait, no, 0.9 - 0.048 = 0.852 (85.2%), 0.9 + 0.048 = 0.948 (94.8%). So the confidence interval is (85.20%, 94.80%).

Wait, but earlier I thought 135/150 was 0.89, but it's 0.9. So that's the mistake. 135 ÷ 150 = 0.9, so \( \hat{p}=0.9 \).

So let's redo:

Step1: Calculate sample proportion: \( \hat{p}=\frac{135}{150}=0.9 \)

Step2: Standard error \( SE=\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}=\sqrt{\frac{0.9\times0.1}{150}}=\sqrt{\frac{0.09}{150}}\approx0.0245 \)

Step3: Margin of error for 95% confidence: \( ME = 1.96\times0.0245\approx0.048 \)

Step4: Confidence interval: \( 0.9\pm0.048 \), which is \( (0.852, 0.948) \) or \( (85.20\%, 94.80\%) \)

So the correct option is \( CI=(85.20\%, 94.80\%) \)

Answer:

CI = (85.20%, 94.80%)