Question

A report on the nightly news broadcast stated that 10 out of 108 households with pet dogs were burglarized and 20 out of 208 without pet dogs were burglarized. Suppose that this data is used to test the claim that the proportion of households with dogs that are burglarized is the same as the proportion of households without dogs that are burglarized. What would be the test statistic for this test

Answer 1

The answer is "0.9184".

Explanation:

This problem is assisted by data analysis:

Its overview hypothesis test to two sample proportions:

`p_1 :`percentage of population contributions 1

`p_2:`percentage of population contributions 2

`\to p_1 - p_2:` proportion difference

`\to H_0 : p_1 - p_2 = 0\\\\\to H_A : p_1 - p_2 ? 0`

Calculating the Hypothesis test values:

`Diff \ \ Count1 \ \ Total1 \ \ Count2 \ \ Total2 \ \ Sam. Diff. \ \ Std. Err.\ \ Z-Stat \ \ P value\\\\`

`p_1 - p_2\ \ 10\ \ 108\ \ 20\ \ 208\ \ -0.0035612536\ \ 0.034766147\ \ -0.10243452 \ \ 0.9184`

So, the
`P - value = 0.9184`