A superintendent of a school district conducted a survey to find out the level of job satisfaction among teachers. Out of 53 teachers who replied to the survey, 13 claim they are satisfied with their job.
z equals fraction numerator p with hat on top minus p over denominator square root of begin display style fraction numerator p q over denominator n end fraction end style end root end fraction
The superintendent wishes to construct a significance test for her data. She find that the proportion of satisfied teachers nationally is 18.4%.
What is the z-statistic for this data? Answer choices are rounded to the hundredths place.
a. 2.90
b. 1.15
c. 1.24
d. 0.61

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joaobezerra joaobezerra · Answer 1 · 2020-07-15T02:25:01+02:00

Answer:

b. 1.15

Step-by-step explanation:

The z statistics is given by:

[tex]Z = \frac{X - p}{s}[/tex]

In which X is the found proportion, p is the expected proportion, and s, which is the standard error is [tex]s = \sqrt{\frac{p(1-p)}{n}}[/tex]

Out of 53 teachers who replied to the survey, 13 claim they are satisfied with their job.

This means that [tex]X = \frac{13}{53} = 0.2453[/tex]

She find that the proportion of satisfied teachers nationally is 18.4%.

This means that [tex]p = 0.184[/tex]

Standard error:

p = 0.184, n = 53.

So

[tex]s = \sqrt{\frac{0.184*0.816}{53}} = 0.0532[/tex]

Z-statistic:

[tex]Z = \frac{X - p}{s}[/tex]

[tex]Z = \frac{0.2453 - 0.184}{0.0532}[/tex]

[tex]Z = 1.15[/tex]

The correct answer is:

b. 1.15