Question

2. A random sample of 284 sixteen and seventeen year olds indicated that 84 of them had sent a text while driving.
   1. What is the 85.02% confidence level of the population proportion of sixteen and seventeen year olds that text while driving. (5)

2. If you wanted the 85.02% confidence interval to be half as wide as what you found in part a., how large would the sample size have to be? (3)

3. Test the hypothesis that the population proportion of sixteen and seventeen year olds that text while driving is greater than 25% at the 5% significance level. To receive full points you must show all of your steps. (5)

Answer 1

n = sample size = 284

x = number of sixteen and seventeen year old that text while driving = 84

Sample proportion:

(Round to 4 decimal)

85.02% confidence level of the population proportion of sixteen and seventeen year old that text while driving is

where zc is z critical value for (1+c)/2 = (1+0.8502)/2 = 0.9251

zc = 1.44 (From statistical table of z values)

(Round to 4 decimal)

85.02% confidence level of the population proportion of sixteen and seventeen year old that text while driving is (0.2568,0.3348)

b)

85.02% Confidence interval is (0.2568,0.3348)

Width of confidence interval is = 0.3348-0.2568 = 0.078

Half of width = 0.078/2 = 0.039

Margin of error (e) = 0.039

Sample size (n) :

n = 283.9814

n = 284 (Round to nearest integer)

Sample size = 284

c)

Here we have to test that

Null hypothesis :

Alternative hypothesis :

where

Test statistic:

z = 1.78 (Round to 2 decimal)

Test statistic = 1.78

Test is right tailed test.

P value = P(z > 1.78)

= 1 - P(z < 1.78)

= 1 - 0.9625 (From statistical table of z values)

= 0.0375

P value = 0.0375

Significance level = = 0.05

Here p value <

So we reject H0.

Conclusion: There is sufficient evidence to support the claim that the population proportion of sixteen and seventeen year old that text while driving is greater than 25% at the 5% significance level.