A survey was conducted in the United Kingdom, where respondents were asked if they had a university degree. One question asked, "In the last 20 years the proportion of the world population living in extreme poverty has ... ", and three choices were provided: 1.) "increased" 2.) "remained more or less the same" and 3.) "decreased". Of 373 university degree holders, 45 responded with the correct answer: decreased; of 639 non-degree respondents, 57 responded with the correct answer1. We would like to test if the percent of correct answers is significantly different between degree holders and non-degree holders.

Let group 1 be the degree holders and let group 2 be the non-degree holders.

a. What are the null and alternative hypotheses?

b. Using technology, construct a randomization distribution and compute the p-value.

Here we have given two catogaries as degree holder and non degree holder.

So here we have to test the hypothesis that,

H0 : p1 = p2 Vs H1 : p1 not = p2

where p1 is population proportion of degree holder.

p2 is population proportion of non degree holder.

Assume alpha = level of significance = 5% = 0.05

The test is two tailed.

Here test statistic follows standard normal distribution.

The test statistic is,

Z = (p1^ - p2^) / SE

where SE = sqrt[ (p^*q^) / n1 + (p^*q^) / n2]

p1^ = x1/n1

p2^ = x2/n2

p^ = (x1+x2) / (n1+n2)

This we can done in TI_83 calculator.

steps:

STAT - -> TESTS - -> 6:2-PropZTest - -> ENTER - -> Input all the values - -> select alternative "not = P2" - -> ENTER - -> Calculate - -> ENTER

Test statistic Z = 1.60

P-value = 0.1090

P-value > alpha

Fail to reject H0 or accept H0 at 5% level of significance.

Conclusion : There is not sufficient evidence to say that the percent of correct answers is significantly different between degree holders and non-degree holders.