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 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.

so, converting firstly the mixed fractions to improper fractions and then "adding".