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How do I solve this STATS question?

Based on a survey of a random sample of 900 adults in the United States, a journalist reports that 60 percent of adults in the United States are in favor of increasing the minimum hourly wage. If the reported percent has a margin of error of 2.7 percentage points, which of the following is closest to the level of confidence?

(A) 80.0%
(B) 90.0%
(C) 95.0%
(D) 95.5%
(E) 99.0%

Respuesta :

We have the following data:

Margin of Error = E = 2.7 % = 0.027
Sample size = n = 900
Proportion of adults in favor = p = 60% = 0.6

We need to find the confidence level. For this first we need to find the z value.

The margin of error for a population proportion is given as:

[tex]E=z* \sqrt{ \frac{p(1-p)}{n} } [/tex]

Using the values, we get:

[tex]0.027=z* \sqrt{ \frac{0.6*0.4}{900} } \\ \\ 0.027=z*0.016 \\ \\ z= \frac{0.027}{0.01633} \\ \\ z=1.65 [/tex]

As, seen from the z table, z=1.65 corresponds to the confidence level 90%. So, the answer to this question is option B
abidemiokin

The equivalent level of confidence for 1.65 is 90% from the z-table

To get the required level of confidence, we will use the formula for calculating the margin of error expressed as:

[tex]E=z\cdot\sqrt{\frac{p(1-p)}{n} }[/tex]

Given the following parameters

Sample size n = 900

p is the ratio = 0.6 * 900 = 540

p = 540/900

p = 0.6

E = 2.7% = 0.027

Substitute the given parameters into the formula

[tex]0.027=z\cdot\sqrt{\frac{0.6(1-0.6)}{900} }\\0.027=z\cdot\sqrt{\frac{0.24}{900} }\\0.027 = 0.0163z\\z = \frac{0.027}{0.0163} \\z =1.65[/tex]

The equivalent level of confidence for 1.65 is 90% from the z-table

Learn more  here: https://brainly.com/question/17042789

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