Question

Perform a one-tailed test. Set up HO and Ha. Compute test statistic and p-value. Draw conclusions. #1. A Gallup survey conducted April 2–11, 2018, found that 152 out of 1015 American adults mentioned immigration/illegal aliens as the most important problem facing the nation. An earlier survey conducted in February 2011 found that 108 out of 978 American adults mentioned immigration/illegal aliens as the most important problem facing the nation. Can we conclude that the percentage of all American adults who believed that immigration/illegal aliens was the most important problem facing the nation was increased in 2018 compared to that in 2011? #2. The District of Columbia Opportunity Scholarship Program was created by Congress to provide tuition vouchers to low-income parents who want their child to attend a private school. In 2012–2014, a treatment group of 995 students were randomly selected to receive a tuition voucher, while a control group of 776 students did not receive vouchers. One year later, the students were given a standardized reading test. The average score of the students in the treatment group was 601.78 with a standard deviation of 52.65; the average score of the control group was 605.78 with a standard deviation of 55.73. At the a = 0.01 level of significance, can we conclude that the average reading score of students who receive a tuition voucher to attend private school is lower than the average reading score of students who do not receive a tuition voucher?

Answer 1

1.

Let the 2018 survey be 1 and 2011 survey be 2. Then,

152 1015 For sample 1, we have that the sample size is Ni 1015, the number of favorable cases is X1 152, so then the sample proportion is fi = x 0.1498 For sample 2, we have that the sample size is N2 = 978, the number of favorable cases is X2 = 108, so then the sample proportion is f2 = x = 0.1104 108 978 The value of the pooled proportion is computed as p X+X, Ni+N2 152+108 = 0.1305 1015+978 Also, the given significance level is a = 0.05. (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: P1 = P2 Ha : Pı > P2

(2) Rejection Region Based on the information provided, the significance level is a = 0.05, and the critical value for a right-tailed test is ze = 1.64. The rejection region for this right-tailed test is R={z:z> 1.64} (3) Test Statistics The Z-statistic is computed as follows: 2 = P1 - P2 p(1-P)(1/n; +1/n2) 0.1498 -0.1104 0.1305 - (1 -0.1305) (1/1015 +1/978) 2.606 (4) Decision about the null hypothesis Since it is observed that z = 2.606 > Zc = 1.64, it is then concluded that the null hypothesis is rejected. Using the P-value approach: The p-value is p=0.0046, and since p=0.0046 < 0.05, it is concluded that the null hypothesis is rejected.

Hence we can say that the percentage of all American adults who believed that immigration/illegal aliens was the most important problem facing the nation was increased in 2018 compared to that in 2011.

2.

Let the group of students who receive a tuition voucher to attend private school be 1 and the other group be 2.

The provided sample means are shown below: X = 601.78 X2 = 605.78 Also, the provided sample standard deviations are: $1 = 52.65 S2 = 55.73 and the sample sizes are ni 995 and n2 = 776 (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: Mi = M2 Ha: Mi M2

(2) Rejection Region Based on the information provided, the significance level is a = 0.05, and the degrees of freedom are df = 1769. In fact, the degrees of freedom are computed as follows, assuming that the population variances are equal: -1.646, for a = 0.05 and df Hence, it is found that the critical value for this left-tailed test is te 1769. The rejection region for this left-tailed test is R= {t:t< -1.646). (3) Test Statistics Since it is assumed that the population variances are equal, the t-statistic is computed as follows: t= X1 - X2 (n-1)s: +(12-1); ni +12-2 (+) 601.78 605.78 -1.546 (995-1)52.652+(776-1)55.732 995-776-2 (995 + 776)

(4) Decision about the null hypothesis Since it is observed that t = -1.546 >te = -1.646, it is then concluded that the null hypothesis is not rejected. Using the P-value approach: The p-value is p = 0.0611, and since p= 0.0611 > 0.05, it is concluded that the null hypothesis is not rejected.

Hence we can't say that  the average reading score of students who receive a tuition voucher to attend private school is lower than the average reading score of students who do not receive a tuition voucher.

Please upvote if you have liked my answer, would be of great help. Thank you.