Mathematics: 5. During a national debate on changes to health care, a cable news service performs an opinion poll of 500 small business owners. It shows that 65% of small-business owners do not approve of health care changes. Develop a 95% confidence interval for the proportion opposing health care changes. Use 4 decimal places.

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5. During a national debate on changes to health

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31.01.2022

The 95% confidence interval for the proportion opposing health care changes is (0.6082, 0.6918).

Step-by-step explanation:

The (1 - α)% confidence interval for the population proportion is:

$CI=\hat p\pm z_{\alpha/2}\cdot\sqrt{\frac{\hat p(1-\hat p)}{n}}$

The information provided is:

$\hat p=0.65\\n=500\\\text{Confidence level}=95\%$

The critical value of z for 95% confidence level is:

$z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96$

*Use a z-table.

Compute the 95% confidence interval for the proportion opposing health care changes as follows:

$CI=\hat p\pm z_{\alpha/2}\cdot\sqrt{\frac{\hat p(1-\hat p)}{n}}$

$=0.65\pm 1.96\sqrt{\frac{0.65(1-0.65)}{500}}\\\\=0.65\pm 0.04181\\\\=(0.60819, 0.69181)\\\\\approx (0.6082, 0.6918)$

Thus, the 95% confidence interval for the proportion opposing health care changes is (0.6082, 0.6918).

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