Question

annie is concerned over a report that a woman over age 40 has a better chance of being killed by a terrorist than of getting married. a study found that the likelihood of marriage for a never-previously-wed, 40 -year-old university-educated american woman was 2.3% . to demonstrate that this percentage is too small, annie uses her resources at the baltimore sun to conduct a simple random sample of 476 never-previously-wed, university-educated, american women who were single at the beginning of their 40 s and who are now 45 . of these women, 16 report now being married. does this evidence support annies claim, at the 0.10 level of significance, that the chances of getting married for this group is greater than 2.3% ? step 2 of 3 : compute the value of the test statistic. round your answer to two decimal places.

Answer 1

The computed value of the test statistic is approximately z=0.05.

To test Annie's claim that the chances of getting married for this group are greater than 2.3%, you can use a one-sample proportion z-test. The null hypothesis (H0) is that the true proportion is 2.3%, and the alternative hypothesis (H1) is that the true proportion is greater than 2.3%.

The formula for the z-test statistic for a one-sample proportion is:

​z = p- p_0/ p_0(1-p_0)/n

Where:

p​ is the sample proportion (16/476 in this case),

p _0 is the hypothesized population proportion under the null hypothesis (2.3% or 0.023),

n is the sample size (476 in this case).

Let's calculate the z-test statistic:

z= 0.023(1−0.023)/ 476

z= 0.033613−0.023/ 0.023×0.977/ 476

z= 0.010613/ 476 ×0.022431

​z≈ 0.010613/0.000047154

z≈ 0.010613/ 0.217136

z≈0.0489

Now, round the test statistic to two decimal places:

z≈0.05

So, the computed value of the test statistic is approximately z=0.05.