A report summarizes data from independent representative samples of approximately 1,000 people in each
of 28 different countries. The report asked people if they thought that, in general, teachers in their country
were trustworthy or untrustworthy. In the sample from the United States, 58% responded trustworthy. In
the sample from Canada, 64% responded trustworthy. For this exercise, assume that the two sample sizes
were 1,000. Based on these data, can you conclude that the proportion of people in the United States who
think that teachers in their country are trustworthy is less than this proportion for people in Canada? (Use
š¼ = 0.05. Let p be the population proportion of people in the United States who think their teachers are
trustworthy and p be the population proportion of people in Canada who think their teachers are
trustworthy.)
a) State the appropriate null and alternative hypotheses.
b) Find the test statistic.
c) Find the p-value
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a) The appropriate null and alternative hypotheses for this test are:
Null Hypothesis (H0): The proportion of people in the United States who think their teachers are trustworthy is equal to or greater than the proportion for people in Canada.
Alternative Hypothesis (HA): The proportion of people in the United States who think their teachers are trustworthy is less than the proportion for people in Canada.
b) The test statistic for this hypothesis test is the z-score.
The test statistic formula for comparing two proportions is:
z = (p1 - p2) / ā(p(1 - p) / n1 + p(1 - p) / n2)
where:
p1 = proportion of people in the United States who think their teachers are trustworthy
p2 = proportion of people in Canada who think their teachers are trustworthy
p = pooled proportion of people who think their teachers are trustworthy (p = (x1 + x2) / (n1 + n2))
n1 = sample size of the United States
n2 = sample size of Canada
In this case, p1 = 0.58, p2 = 0.64, n1 = n2 = 1000.
First, calculate the pooled proportion:
p = (0.58 * 1000 + 0.64 * 1000) / (1000 + 1000) = 0.61
Next, calculate the standard error (SE):
SE = ā(p(1 - p) / n1 + p(1 - p) / n2) = ā(0.61(1 - 0.61) / 1000 + 0.61(1 - 0.61) / 1000) = 0.0134
Finally, calculate the z-score:
z = (0.58 - 0.64) / 0.0134 = -4.48
c) To find the p-value, we need to determine the probability of observing a z-score as extreme or more extreme than -4.48 under the null hypothesis.
Using a standard normal distribution table or calculator, we find that the p-value is less than 0.0001.
Therefore, the p-value is less than the significance level of 0.05, suggesting strong evidence against the null hypothesis. We can conclude that the proportion of people in the United States who think that teachers in their country are trustworthy is less than the proportion for people in Canada.
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