Question

1) A number of initiatives on the topic of legalized gambling have appeared on state ballots. A political candidate has decided to support legalization of casino gambling if he is convinced that more than two-thirds of U.S. adults approve of casino gambling. Suppose that 1527 adults selected at random from households with telephones were asked whether they approved of casino gambling. The number in the sample who approved was 1032. Does the sample provide convincing evidence that more than two-thirds of U.S. adults approve? (Use α = 0.05.)

Round your test statistic to two decimal places and your P-value to four decimal places.

z= _

p- value= _

Answer 1

Here, we have to use one sample z test for the population proportion.

The null and alternative hypotheses for this test are given as below:

H0: p = 2/3 versus Ha: p > 2/3

This is an upper tailed test.

We are given

Level of significance = α = 0.05

Test statistic formula for this test is given as below:

Z = (p̂ - p)/sqrt(pq/n)

Where, p̂ = Sample proportion, p is population proportion, q = 1 - p, and n is sample size

x = number of items of interest = 1032

n = sample size = 1527

p̂ = x/n = 1032/1527 = 0.6758

p = 2/3 = 0.6667

q = 1 - p = 0.3333

Z = (p̂ - p)/sqrt(pq/n)

Z = (0.6758 - 0.6667)/sqrt(0.6667*0.3333/1527)

Z = 0.7573

Test statistic = 0.7573

P-value = 0.2244

(by using z-table)

P-value > α = 0.05

So, we do not reject the null hypothesis

The sample does not provide the convincing evidence that more than two-thirds of U.S. adults approve.