Question

28. is smoking more common in Nova Scotia than in Ontario? In a random sample of 140 adults from Nova Scotia there were 40 smokers. The proportion of adult smokers in Ontario is estimated to be 20%

A) at the 0.05 significance level, test if the smoking is more common in Nova Scotia

B) Find the p-value associated with this test

C) Suppose the test statistic is 2.20 can we conclude that Hn should be rejected at 1) X-0.01 level of Type 1 error

Answer 1

A) H₀: p = 0.2

    H₁: p > 0.2

$\widehat p$ = 40/140 = 0.2857

The test statistic z = ( $\widehat p$ - p)/sqrt(p(1 - p)/n)

                             = (0.2857 - 0.2)/sqrt(0.2 * 0.8/140)

                            = 2.54

At 0.05 significance level, the critical value is z_0.95 = 1.645

Since the test statistic value is greater than the critical value(2.54 > 1.645), so we should reject the null hypothesis.

So at 0.05 significance level there is sufficient evidence to conclude that the smoking is more common in Nova Scotia.

B) P-value = P(Z > 2.54)

                  = 1 - P(Z < 2.54)

                  = 1 - 0.9945

                  = 0.0055

C) P-value = P(Z > 2.20)

                 = 1 - P(Z < 2.20)

                = 1 - 0.9861

                = 0.0139

At 0.01 significance level since the P-value is greater than the significance level(0.0139 > 0.01), so we should reject the null hypothesis.