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A random sample of 350 voters from Delaware finds that 105 of them intend to vote for Bill McNeely in an upcoming election for Governor. Construct a 96% confidence interval for the population proportion of voters who intend to vote for Bill McNeely. Enter the lower and upper bounds for the interval in the following boxes, respectively. You may answer using decimals rounded to four places or a percentage rounded to two. Make sure to use a percent sign if you answer using a percentage.

Answer :

Answer:

Lower bound: 0.2497 = 24.97%

Upper bound: 0.3503 = 35.03%

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of [tex]\pi[/tex], and a confidence level of [tex]1-\alpha[/tex], we have the following confidence interval of proportions.

[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

In which

z is the zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex].

For this problem, we have that:

A random sample of 350 voters from Delaware finds that 105 of them intend to vote for Bill McNeely in an upcoming election for Governor. This means that [tex]n = 350, p = \frac{105}{350} = 0.3[/tex]

96% confidence level

So [tex]\alpha = 0.04[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.04}{2} = 0.98[/tex], so [tex]Z = 2.055[/tex].

The lower limit of this interval is:

[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.3 - 2.055\sqrt{\frac{0.3*0.7}{350}} = 0.2497[/tex]

The upper limit of this interval is:

[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.3 + 2.055\sqrt{\frac{0.3*0.7}{350}} = 0.3503[/tex]

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