Mathematics, 09.04.2020 18:58 kelseeygee

Fix the roads~The Township Board of Meridian Township wants to know how much public support there is for raising property taxes to fix and maintain the roads in the township. They randomly surveyed 120 residents and calculated a 90% confidence interval for the actual proportion of Meridian Township residents who are in favor of the tax increase to be (0.526, 0.674). Which of the following is a 99% confidence interval using the same survey data?A. (.512, .688)B. (.48, .715)C. (.536, .664)

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Answer from: cacaface311

The 99% confidence interval of the sample proportion obtained for the 95% confidence interval is (0.48; 0.715)

From the interval given :

Sample proportion, p = (0.526+0.674) / 2 = 0.60

Standard deviation of proportion :

$\sqrt{\frac{p(1 - p)}{n}}$

Hence, the standard deviation, σ is :

$\sqrt{\frac{0.60(0.40)}{120}} = 0.0447$

Z* = The Critical value at 99% confidence interval = 2.576

The confidence interval can be calculated thus :

p ± Z*σ

Lower boundary = (0.60 - 2.576(0.0447)) = 0.4848

Upper boundary = (0.60 - 2.576(0.0447)) = 0.715

Therefore, the confidence interval is (0.485 ; 0.715)

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Answer from: aidanfbussiness

B) 99% confidence interval is (0.485,0.715)

Step-by-step explanation:

We are given a 90% confidence interval constructed for a population proportion using a sample size of 120. We have to construct a 99% confidence interval from this data. For this first we need the value of sample proportion i.e. the point estimate.

The 90% confidence interval is: (0.526, 0.674). Remember that the point estimate is at an equal distance from both upper limit and lower limit of the confidence interval. So finding out the average of both the limits will give us their middle value i.e. the point estimate or the value of sample proportion.

So, the value of sample proportion will be:

$p=\frac{0.526+0.674}{2}=0.60$

Now we can calculate the 99% confidence interval using the formula:

$(p-z_{\frac{\alpha}{2} } \times \sqrt{\frac{p(1-p)}{n} }, p-z_{\frac{\alpha}{2} } \times \sqrt{\frac{p(1-p)}{n} })$

Here, $z_{\frac{\alpha}{2}$ is the critical z value for given confidence level. For 99% confidence level, the z value from z-table comes out to be 2.576. The sample size n is given to be 120. Using all these values, we get:

$(0.60-2.576 \times \sqrt{\frac{0.60(1-0.60)}{120}}, 0.60+2.576 \times \sqrt{\frac{0.60(1-0.60)}{120}})\\\\ =(0.485,0.715)$

Therefore, 99% confidence interval is (0.485,0.715)

Answer from: Quest

the sum of the lengths of any two sides of a triangle is greater than the length of the third side. if you take the three sides of a triangle and add them in pairs, the sum is greater than (not equal to) the third side. if that is not true, then it is not possible to construct a triangle with the given side lengths.

step-by-step explanation:

Which are the possible side lengths of a triangle?

Answer from: Quest

step-by-step explanation:

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