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azamat
3 years ago
5

Find the value of z Subscript alpha divided by 2 that corresponds to a confidence level of​ 89.48%.

Mathematics
1 answer:
Katyanochek1 [597]3 years ago
4 0

Answer:

For this case we know that the confidence level is 89.48% or 0.8948 and the significance would be: \alpha=1-0.8948 = 0.1052 and the value of \alpha/2 =0.0526

Now we need to find a quantile in the normal standard distribution who accumulate 0.0526 of the area on each tail of the normal standard distribution and we got:

z_{\alpha/2}= \pm 1.62015

And we can use the following excel code for example:

"=NORM.INV(0.0526,0,1)"

Step-by-step explanation:

For this case we know that the confidence level is 89.48% or 0.8948 and the significance would be: \alpha=1-0.8948 = 0.1052 and the value of \alpha/2 =0.0526

Now we need to find a quantile in the normal standard distribution who accumulate 0.0526 of the area on each tail of the normal standard distribution and we got:

z_{\alpha/2}= \pm 1.62015

And we can use the following excel code for example:

"=NORM.INV(0.0526,0,1)"

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Step-by-step explanation:

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3 years ago
Suppose that a local TV station conducts a survey of a random sample of 120 registered voters in order to predict the winner of
forsale [732]

Answer:

a) The 99% CI for the true proportion of voters who prefer the Republican candidate is (0.3658, 0.6001). This means that we are 99% sure that the true population proportion of all voters who prefer the Republican candidate is (0.3658, 0.6001).

b) The upper bound of the confidence interval is above 0.5 = 50%, which meas that the candidate can be confidence of victory.

Step-by-step explanation:

Question a:

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

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

z is the z-score that has a p-value of 1 - \frac{\alpha}{2}.

Sample of 120 registered voters in order to predict the winner of a local election. The Democrat candidate was favored by 62 of the respondents.

So 120 - 62 = 58 favored the Republican candidate, so:

n = 120, \pi = \frac{58}{120} = 0.4833

99% confidence level

So \alpha = 0.01, z is the value of Z that has a p-value of 1 - \frac{0.01}{2} = 0.995, so Z = 2.575.  

The lower limit of this interval is:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.4833 - 2.575\sqrt{\frac{0.4833*0.5167}{120}} = 0.3658

The upper limit of this interval is:

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.4833 + 2.575\sqrt{\frac{0.4833*0.5167}{120}} = 0.6001

The 99% CI for the true proportion of voters who prefer the Republican candidate is (0.3658, 0.6001). This means that we are 99% sure that the true population proportion of all voters who prefer the Republican candidate is (0.3658, 0.6001).

b. If a candidate needs a simple majority of the votes to win the election, can the Republican candidate be confident of victory? Justify your response with an appropriate statistical argument.

The upper bound of the confidence interval is above 0.5 = 50%, which meas that the candidate can be confidence of victory.

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3 years ago
The land area of Charlotte, NC is approximately 3 x 102 square miles. There are 6.4 x 102 square acres in a square mile. How man
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Step-by-step explanation:

Divide 6.4 acres by 3 miles = factor of 2.13

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