Question

Use the information to complete the following questions. Show organized set-ups.

The Arizona Republic conducted a telephone poll of 200 Arizona adults who celebrated Christmas. The question asked was, “In your family, do you open presents on Christmas Day?” Of those surveyed, 52% said they do open presents on Christmas Day.

1. Construct a 95% confidence interval for the true proportion of Arizona adults who celebrate Christmas that open presents on Christmas Day. (6 pt)

2. You want to see if the proportion of Iowans that use E85 fuel in their vehicle is different from the national average of 51%. You survey 378 Iowans and 223 said they use E85 fuel in their vehicle. Complete the following:

-Write the null and alternative hypothesis for this situation. (4 pt)

-Calculate the test statistic. (6 pt)

-Find the p-value (hint-sketch the normal curve also) (3 pt)

-Using a significance level of 0.05, write a conclusion for the hypothesis test. Be sure to include the topic and hypothesis (Don’t just write “reject”, “do not reject”) (4 pt)

-Define p-value (3 pts)

*Note: Please show work for all parts. Thank you!

Answer 1

1) At 95% confidence interval the critical value is z_0.025 = 1.96

The 95% confidence interval is

+/- z_0.025 * sqrt((1 - )/n)

= 0.52 +/- 1.96 * sqrt(0.52 * (1 - 0.52)/200)

= 0.52 +/- 0.069

= 0.451, 0.589

2) H₀: P = 0.51

    H₁: P 0.51

= 223/378 = 0.59

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

                             = (0.59 - 0.51)/sqrt(0.51 * (1 - 0.51)/378)

                             = 3.11

P-value = 2 * P(Z > 3.11)

             = 2 * (1 - P(Z < 3.11))

             = 2 * (1 - 0.9991)

             = 2 * 0.0009 = 0.0018

Since the P-value is less than the significance level(0.0018 < 0.05), so we should reject the null hypothesis.

At 0.05 significance level, there is sufficient evidence to conclude that the proportion of lowans that use E85 fuel in their vehicle is different from the national average of 51%.