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Answer:
The confidence interval for the mean is given by the following formula: Â
[tex]\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}[/tex]
And the margin of error is given by:
[tex] ME = 1.96 *\frac{0.87*(1-0.87)}{100} = 0.0659[/tex]
Step-by-step explanation:
Previous concepts
A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval". Â
The margin of error is the range of values below and above the sample statistic in a confidence interval. Â
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean". Â
Solution to the problem
In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by [tex]\alpha=1-0.95=0.05[/tex] and [tex]\alpha/2 =0.025[/tex]. And the critical value would be given by:
[tex]z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96[/tex]
The confidence interval for the mean is given by the following formula: Â
[tex]\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}[/tex]
And the margin of error is given by:
[tex] ME = 1.96 *\frac{0.87*(1-0.87)}{100} = 0.0659[/tex]
The margin of error at 95% confidence interval is 0.0659
Data;
- n = 100
- confidence level = 95%
- probability = 85%
Margin of Error
To calculate the margin of error, we can use a formula to solve this.
[tex]E = Z_\frac{\alpha }{2} \sqrt{\frac{p(1-p)}{n} }\\E = 0.066[/tex]
At 95% confidence level, we can substitute the value and solve.
[tex]E = 1.960 * \sqrt{\frac{0.87*0.13}{100} }\\E = 0.0659[/tex]
The margin of error at 95% confidence interval is 0.0659
Learn more on margin of error here;
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