In recent survey, a school district randomly dialed 300 parent phone numbers and asked whether the family read to their toddlers at least three nights per week. 271 of the parents said that they did.
(a) Nationally , 72% of families read to their toddlers at least three nights per week. Does this data provide evidence that more parents at this district read to their children than the national average ? Show all steps in your process.
(b) Name two likely sources of bias in this survey. Write one sentence to explain each.

b) 1.-The survey was made over a district, results will surely be different if the survey is carried out over a whole state ( considering urban and rural areas)

2.-In a district we find an equalized level of salaries, which could be associated with a similar level of habits

Step-by-step explanation:

We have to develop a proportion test. One tail-test

Population proportion mean (national proportion) p₀ = 72 %

Sample information

sample proportion 271/300 p = 0,90 p = 90 %

We assume Confidence Interval 90 % then α = 0,1

Test Hypothesis

Null Hypothesis H₀ p = p₀

Alternative Hypothesis Hₐ p > p₀

As α = 0,1 and

We look at z-table for z(c) and find z(c) = 1,28

And we compute z(s) as

z(s) = ( p - p₀ ) √ (p₀q₀/n

z(s) = ( 0,9 - 0,72 ) /√(0,72)*(0,28)/300

z(s) = 0,18 / √0,1296/300

z(s) = 0,18/ 0,026

z(s) = 6,92

z(s) > z(c) 6,92 > 1,28

As z(s) is bigger than z(c), z(s) is in the rejection region, so we reject the null hypothesis. We have enough evidence to claim that the proportion in the district is bigger than the national one.

b)1. The survey was made over a district, results will surely be different if the survey is carried out over a whole state ( considering urban and rural areas)

2.-In a district we find an equalized level of salaries, which could be associated with a similar level of habits