Question

It has been found in previous studies that 70% of all teenage smokers end up quitting. In order to test this statistic, a sample of 87 teenage smoker were followed for twenty-five years to determine if they continued smoking or not. Out of the 87 teenagers surveyed, 68 ended up quitting. At the .05 los, can we conclude that more than 70% of teenage smokers ended up quitting?

a) State the null and alternate hypothesis:

b) Name the Test: ________________________
c) Show that the conditions needed to perform this test are satisfied:

d) Determine the P-value: ___________________________ Write as a decimal.
e) Determine whether to Reject H0 or Fail to Reject H0 , and TELL WHY by relating P-value to α:

f) Interpret the result in a complete sentence with wording related to the problem:

g) How large a sample (find n) is needed in order to be 90% sure that is within a distance of 0.10 from p? Show work with correct formula. Leave answer as a whole person.

Answer 1

a) H₀: P = 0.7

H₁: P > 0.7

b) One sample test of proportion.

c) np(1 - p) = 87 * 0.7 * 0.3 = 18.27 is greater than 10 and the sample size is less than Or equal to 5% of the population size and the conditions are meet to perform the hypothesis test.

d) = 68/87 = 0.7816

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

= (0.7816 - 0.7)/sqrt(0.7 * 0.3/87)

= 1.66

P-value = P(Z > 1.66)

= 1 - P(Z < 1.66)

= 1 - 0.9515

= 0.0485

e) Since the p-value is less than = 0.05, so we should reject the null hypothesis.

f) At 5% level of significance, we can conclude that more than 70% of teenage smokers ended up quitting.

g) At 90% confidence interval the critical value is z_0.05 = 1.645

Margin of error = 0.10

Or, z_0.05 * sqrt(p(1 - p)/n) = 0.10

Or, 1.645 * sqrt(0.7 * 0.3/n) = 0.10

Or, n = (1.645 * sqrt(0.7 * 0.3)/0.10)^2

Or, n = 57