Question

15. A researcher wants to know the poverty rate for the state of Kansas. The researcher randomly selects 150 families and finds 25 are at or below the poverty line.

a) Determine a 95% confidence interval for the proportion. State this interval below within an interpretive sentence tied to the given context.

b) The poverty rate for the US is stated to be 12.3%. Is there evidence at the 95% confidence level that the the poverty rate is higher than 12.3%? Explain your reasoning.

Answer 1

Solution :

Given that,

n = 150

x = 25

Point estimate = sample proportion = = x / n = 25 / 150=0.167

1 - = 1 - 0.167=0.833

At 95% confidence level

= 1 - 95%

=1 - 0.95 =0.05

/2 = 0.025

Z/2 = Z_{0.025 = 1.96}

Margin of error = E = Z _{/ 2} * (( * (1 - )) / n)

= 1.96 (((0.167*0.833) /150 )

= 0.0597

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.167 - 0.0597 < p < 0.167 + 0.0597

0.1073< p < 0.2267

The 95% confidence interval for the population proportion p is :(0.1073 ,  0.2267)