Question

it is thought that not as many americans buy presents to celebrate valentines day anymore. a random sample of 4000 americans yielded 2200 who bought their significant other a present and celebrated valentines day. estimate the true proportion of all americans who celebrate valentines day using a 98% confidence interval. express the answer in the form $hat{p}pm e$ and round to four decimal places.
oa. $0.4375pm0.0183$
ob. $0.5500pm0.5727$
oc. $0.4375pm0.5727$
od. $0.5500pm0.0183$

Explanation:

Step1: Calculate sample proportion \(\hat{p}\)

The sample proportion \(\hat{p}=\frac{x}{n}\), where \(x = 2200\) (number of successes) and \(n=4000\) (sample size). So \(\hat{p}=\frac{2200}{4000}=0.55\)

Step2: Find \(z\) - value for \(98\%\) confidence interval

For a \(98\%\) confidence interval, the significance level \(\alpha=1 - 0.98=0.02\), and \(\alpha/2=0.01\). Using the standard normal distribution \(z_{\alpha/2}=z_{0.01}\approx 2.33\)

Step3: Calculate the margin of error \(E\)

The formula for the margin of error for a proportion is \(E = z_{\alpha/2}\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}\)
Substitute \(\hat{p}=0.55\), \(n = 4000\), and \(z_{\alpha/2}=2.33\)

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Answer:

D. \(0.5500\pm0.0183\)