Help with a statistics problem of proving the view, given a 2-tail and an α=.01?

Question

Compare the percentage of people who are afraid of walking in their neighborhood at night across the TV viewing variable (2-tail, α=.01) using the following information:

 

Proportion Afraid in Neighborhood by TV Viewing

Hours of TV:

====================

__(<)Avg_____(≥)Avg.

p:__.32________.29

n:___442______188

 

How would one interpret the results in the context of the view that TV viewing leads to less civic engagement which, in turn leads to greater fear of crime. Is there support for that view?

Thank you

Answer 1

Consider a two sample proportion test. \begin{align*} \hat{p}&=\frac{n_1\hat{p_1}+n_2\hat{p_2}}{n_1+n_2}\approx0.31\\ Z&=\frac{\hat{p_1}-\hat{p_2}-0}{\sqrt{\hat{p}(1-\hat{p})\left(\frac{1}{n_1}+\frac{1}{n_2}\right)}}=\frac{0.32-0.29}{\sqrt{0.31\times0.69\times\left(\frac{1}{442}+\frac{1}{188}\right)}}\approx0.75<1 \end{align*} So it's not significant. In order for $\alpha=0.01$ significance, we need a Z value much bigger (find exact value from normal table).

Answer 2

Something is strange about the data: $.32(442) = 141.44$ and $.29(188) = 54.52.$ I would have expected something close to an integer in each case.

There is a controversy whether it is best to use a combined estimate of the standard error (in the denominator of $Z$) or whether to estimate the variances separately. Usually, it makes little difference. @Jack has done one version of the test (+1), so I will show the other. You should use whatever formula your text or notes recommend.

Here is Minitab output for the 'separate' method and using 141 and 54 as the counts of 'fearful' subjects. Note that for this $2 \times 2$ table, use of Fisher's exact test is another possibility.

Test and CI for Two Proportions 

Sample    X    N  Sample p
1       141  442  0.319005
2        54  188  0.287234

Difference = p (1) - p (2)
Estimate for difference:  0.0317705
95% CI for difference:  (-0.0461486, 0.109690)
Test for difference = 0 (vs ≠ 0):  Z = 0.80  P-Value = 0.424

Fisher’s exact test: P-Value = 0.452

Neither test shows anything close to a significant result. This is not surprising, the sample fractions of fearful subjects are roughly the same and the sample sizes are small. The 95% margin of sampling error in a public opinion poll of 2500 people is about $\pm 2\%;$ with 1100 people about $\pm 3\%;$ and with 625 people about $\pm 4\%.$