Question

A study considered the question, "Are you a registered voter?" Accuracy of response was confirmed by a check of city voting records. Two methods of survey were used: a face-to-face interview and a telephone interview. A random sample of 93 people were asked the voter registration question face to face. Of those sampled, seventy-eight respondents gave accurate answers (as verified by city records). Another random sample of 83 people were asked the same question during a telephone interview. Of those sampled, seventy-three respondents gave accurate answers. Assume the samples are representative of the general population. (a) Let p1 be the population proportion of all people who answer the voter registration question accurately during a face-to-face interview. Let p2 be the population proportion of all people who answer the question accurately during a telephone interview. Find a 95% confidence interval for p1 – p2. (Round your answers to three decimal places.) lower limit upper limit (b) Does the interval contain numbers that are all positive? all negative? mixed? Comment on the meaning of the confidence interval in the context of this problem. At the 95% level, do you detect any difference in the proportion of accurate responses from face-to-face interviews compared with the proportion of accurate responses from telephone interviews? Because the interval contains only positive numbers, we can say that there is a higher proportion of accurate responses in face-to-face interviews. Because the interval contains both positive and negative numbers, we can not say that there is a higher proportion of accurate responses in face-to-face interviews. We can not make any conclusions using this confidence interval. Because the interval contains only negative numbers, we can say that there is a higher proportion of accurate responses in telephone interviews. . (c) Test the claim that there is a difference in the proportion of accurate responses from face-to-face interviews compared with telephone interviews. Use α = 0.05. (i) What is the level of significance? State the null and alternate hypotheses. H0: p1 = p2; H1: p1 < p2 H0: p1 = p2; H1: p1 > p2 H0: p1 > p2; H1: p1 = p2 H0: p1 = p2; H1: p1 ≠ p2 . (ii) What sampling distribution will you use? What assumptions are you making? The Student's t. We assume that both population distributions are approximately normal with known standard deviations. The Student's t. We assume that both population distributions are approximately normal with unknown standard deviations. The standard normal. We assume that both population distributions are approximately normal with known standard deviations. The standard normal. We assume that both population distributions are approximately normal with unknown standard deviations. What is the value of the sample test statistic? Compute the corresponding z or t value as appropriate. (Test the difference p1 − p2. Round your answer to two decimal places.) (iii) Find (or estimate) the P-value. (Round your answer to four decimal places.) Sketch the sampling distribution and show the area corresponding to the P-value. WebAssign Plot WebAssign Plot WebAssign Plot WebAssign Plot. (iv) Based on your answers to parts (i)-(iii), will you reject or fail to reject the null hypothesis? Are the data statistically significant at level α? At the α = 0.05 level, we reject the null hypothesis and conclude the data are not statistically significant. At the α = 0.05 level, we fail to reject the null hypothesis and conclude the data are not statistically significant. At the α = 0.05 level, we fail to reject the null hypothesis and conclude the data are statistically significant. At the α = 0.05 level, we reject the null hypothesis and conclude the data are statistically significant. (v) Interpret your conclusion in the context of the application. Fail to reject the null hypothesis, there is sufficient evidence to suggest that the proportion of accurate responses from face-to-face interviews differs from the proportion for telephone interviews. Reject the null hypothesis, there is sufficient evidence to suggest that the proportion of accurate responses from face-to-face interviews differs from the proportion for telephone interviews. Reject the null hypothesis, there is insufficient evidence to suggest that the proportion of accurate responses from face-to-face interviews differs from the proportion for telephone interviews. Fail to reject the null hypothesis, there is insufficient evidence to suggest that the proportion of accurate responses from face-to-face interviews differs from the proportion for telephone interviews.

Answer 1

a)
Here, , n1 = 93 , n2 = 83
p1cap = 0.8387 , p2cap = 0.8795

Standard Error, sigma(p1cap - p2cap),
SE = sqrt(p1cap * (1-p1cap)/n1 + p2cap * (1-p2cap)/n2)
SE = sqrt(0.8387 * (1-0.8387)/93 + 0.8795*(1-0.8795)/83)
SE = 0.0523

For 0.95 CI, z-value = 1.96
Confidence Interval,
CI = (p1cap - p2cap - z*SE, p1cap - p2cap + z*SE)
CI = (0.8387 - 0.8795 - 1.96*0.0523, 0.8387 - 0.8795 + 1.96*0.0523)
CI = (-0.143 , 0.062)

b)

Because the interval contains both positive and negative numbers, we can not say that there is a higher proportion of accurate responses in face-to-face interviews.

c)

1)

0.05

H0: p1 = p2; H1: p1 ≠ p2

2)

We assume that both population distributions are approximately normal with known standard deviations.

3)

p1cap = X1/N1 = 78/93 = 0.8387
p1cap = X2/N2 = 73/83 = 0.8795
pcap = (X1 + X2)/(N1 + N2) = (78+73)/(93+83) = 0.858

Test statistic
z = (p1cap - p2cap)/sqrt(pcap * (1-pcap) * (1/N1 + 1/N2))
z = (0.8387-0.8795)/sqrt(0.858*(1-0.858)*(1/93 + 1/83))
z = -0.77

P-value Approach
P-value = 0.4413

4)
At the α = 0.05 level, we fail to reject the null hypothesis and conclude the data are not statistically significant.

5)

Fail to reject the null hypothesis, there is insufficient evidence to suggest that the proportion of accurate responses from face-to-face interviews differs from the proportion for telephone interviews.