Test the hypothesis that public opinion did not change between 1982 and 1994?

Question

A survey asked, "Do you favor or oppose the death penalty for persons convicted of murder?". In 1982, 154 were in favor and 165 opposed. In 1994, 2215 were in favor and 580 opposed. Test the hypothesis that public opinion did not change between 1982 and 1994.

This is a statistical question. Help would be much appreciated. Thank you.

Answer 1

Assuming a random sample, what percentage of total respondents were against the death penalty in 1982 and what percentage in 1994.  If the percentages are the same then you can assume that there was no change in public opinion.

Answer 2

Let PA be the population proportion of people in the public who is in favor the death penalty for persons convicted of murder in 1994. Then let PB to denote the population proportion of people in the public who is in favor of it in 1982.

We wish to set up the following pair of hypothesis to be testd as follows:

We wish to test

H0: PA - PB = 0.

versus

HA:PA- PB ≠ 0.

We are given the following perliminary information: nA= (2215+580)=2795, ^pA = 2215/2795 = 443/559 = 0.792486583, ^SPA=√[^PA*(1-^PA)/nA] = √[0.792486583*(1-0.792486583)/2795] = 0.007670578942 , nB= (154+165)=319, ^pB = 154/319 = 0.48275862, ^SPB = =√[^PB*(1-^PB)/nB] = √[0.48275862*(1-0.48275862)/319] = 0.0027977976

We set to test the above hypothesis at the ∞ = 0.05 level of significance.

Since this test is two-sided, so ∞/2 = 0.05/2 = 0.025.

Z0.025 = 1.96

Since nA*^PA=2795*0.792486583 =2215 and nA*(1-^PA)=2795*(1-0.792486583) = 580 & nB*^PB= 319*0.48275862= 154 and nB*(1-^PB)= 319* (1-0.48275862) = 165 are both at least greater than 5, then nA and nB are considered to be large and their sampling distributions would be normally distributed .Thus it is appropriate to use the z distribution in this case.

We can only reject H0 in favor of Ha if and only if z>z(∞/2).

The test static = z

= [(^PA-^PB)- D0]/√[^PA*(1-^PA)/(nA-1) + ^PB*(1-^PB)/(nB-1)]

= [(0.792486583-0.48275862)- 0]/√[0.792486583*(1-0.792486583)/(2795-1... + 0.48275862*(1-0.48275862)/(319-1)]

= 10.66072314

|Z| = |10.66072314| = 10.66072314

Because |z|= 10.66072314 is much greater than z0.025 = 1.96, then we can reject H0 in favor of Ha at the 0.05 level of significance. The p-value for the test is the sum of the area under the standard normal curve to the right and left of z= and z=respectively. This means the p-value = 2* (0.5-0.4990) = 2*0.001 = 0.002. In other words, we have very strong evidence that there is indeed a difference between the population proportion of people in the public who is in favor of the death penalty for persons convicted of murder in 1982 and 1994 and this means the public opinion has changed. between 1982 and 1994.

Hope it helps.