Question

In 2014 a survey was conducted and found that 297 out of 1100 individuals had confidence in public schools. In 2016 another survey was conducted which led to the result that 322 out of 1150 individuals had confidence in public schools. Let  be the proportion of individuals in 2014 who had confidence in public schools and let  be the proportion of individuals in 2016 who had confidence in public schools.

1.Compute

2.Compute

3.Set up a hypothesis to test in order to decide if the proportion of individuals who have confidence in public schools increased from 2014 to 2016.

4.What is the value of the statistic? Give answer rounded to two decimal places.

Answer 1

This is a comparison between two population proportions. Since n > 30, we can use the normal approximation.

The sample individual proportion and pooled proportion is given below.

x: no. of individual having confidence

n: total individuals sampled

  

We want to test whether the proportion has increased from 2014 to 2016. So if diff = 2014 - 2016 then 2014 < 2016 is to be test 2014 - 2016 <0 . This is left one tailed test.

3.Set up a hypothesis to test in order to decide if the proportion of individuals who have confidence in public schools increased from 2014 to 2016.

(p₁ = p₂)

(p₁ < p₂)

Test Stat=

p-value = P( Z > |Test Stat|)

=1 -P( Z < |Test Stat|) ..................using normal distribution tables

4.What is the value of the statistic? Give answer rounded to two decimal places.

		2014 (1)	2016 (2)
	x	297	322
	n	1100	1150
1.	Sample p	0.27	0.28
	pooled p	0.2751
	SE	0.0188
	Sample diff	-0.0100
	Test Stat	-0.5310
	p-value	0.2981		P(Z < 0.54) =0.7019
	Since p-value > 0.05		level of significnce = 0.05
	Decision	Do not reject the null hypo
	Conclusion	There is insufficient evidence to conclude that the