Question

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

Answer 1

Answer:

b. 1.15

Step-by-step explanation:

The z statistics is given by:

$Z = \frac{X - p}{s}$

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

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

This means that $X = \frac{13}{53} = 0.2453$

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

This means that $p = 0.184$

Standard error:

p = 0.184, n = 53.

So

$s = \sqrt{\frac{0.184*0.816}{53}} = 0.0532$

Z-statistic:

$Z = \frac{X - p}{s}$

$Z = \frac{0.2453 - 0.184}{0.0532}$

$Z = 1.15$

The correct answer is:

b. 1.15