Question

In a recent survey of 60 randomly selected college students, 43 said that they believe in the existence of extraterrestrial life. a) Find p?, the sample proportion that believes that there is extraterrestrial life. (Round your answers to three decimal places). p? = b)The 99 % margin of error associated with this estimate is: c) The 99 % confidence interval for the true proportion of all college students who believe there is extraterrestrial life is: to d) A recent report suggests the proportion of general US population who believe in extraterrestrial life is 0.47. Choose the appropriate null and alternative hypothesis that tests whether the proportion of college students who believe in extraterrestrial life in greater than the general population? H0: p = 0.47 Ha: p ? 0.47 H0: x = 0.47 Ha: x > 0.47 H0: p = 0.47 Ha: p > 0.47 H0: p? = 0.47 Ha: p? > 0.47 H0: ? = 0.47 Ha: ? ? 0.47 e) Calculate the z test statistic and p-value. (Round the test statistic to two decimal places and the p-value four decimal places). z = p-value = f) Based on the p-value, give a conclusion in terms of the alternative. There is suggestive, but inconclusive evidence that the proportion of college students who believe in extraterrestrial life is greater than 0.47. There is convincing evidence that the proportion of college students who believe in extraterrestrial life is greater than 0.47. There is moderately suggestive evidence that the proportion of college students who believe in extraterrestrial life is greater than 0.47. There is no evidence that the proportion of college students who believe in extraterrestrial life is greater than 0.47.

Answer 1

a) = 43/60 = 0.717

b) At 99% confidence interval the critical value is z_0.005 = 2.58

Margin of error = z_0.005 * sqrt((1 - )/n)

                         = 2.58 * sqrt(0.717 * (1 - 0.717)/60)

                          = 0.15

c) The 99% confidence interval is

+/- ME

= 0.717 +/- 0.15

= 0.567, 0.867

d) H₀: P = 0.47

    H_a: P > 0.47

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

                                   = (0.717 - 0.47)/sqrt(0.47(1 - 0.47)/60)

                                   = 3.83

P-value = P(Z > 3.83)

             = 1 - P(Z < 3.83)

             = 1 - 1 = 0

f) For 0.05 significance level, the p-value is less than the significance level (0 < 0.05), so we should not reject the null hypothesis.

There is convincing evidence that the proportion of college students who believe in extraterrestrial life is greater than 0.47.