Question

Although older Americans are most afraid of crime, it is young people who are more likely to be the actual victims of crime. It seems that older people are more cautious about the people with whom they associate. A national survey showed that 10% of all people ages 16-19 have been victims of crime.† At Jefferson High School, a random sample of n = 66 students (ages 16-19) showed that r = 9 had been victims of a crime. Note: For degrees of freedom d.f. not in the Student's t table, use the closest d.f. that is smaller. In some situations, this choice of d.f. may increase the P-value a small amount and thereby produce a slightly more "conservative" answer. (a) Do these data indicate that the population proportion of students in this school (ages 16-19) who have been victims of a crime is different (either way) from the national rate for this age group? Use α = 0.05. Do you think the conditions np > 5 and nq > 5 are satisfied in this setting? Why is this important? (i) What is the level of significance? State the null and alternate hypotheses. H0: μ = 0.10; H1: μ ≠ 0.10 H0: p = 0.10; H1: p ≠ 0.10 H0: μ = 0.10; H1: μ > 0.10 H0: p = 0.10; H1: p > 0.10 H0: p = 0.10; H1: p < 0.10 H0: μ = 0.10; H1: μ < 0.10 (ii) What sampling distribution will you use? What assumptions are you making? The Student's t, since np < 5 and nq < 5. The Student's t, since np > 5 and nq > 5. The standard normal, since np > 5 and nq > 5. The standard normal, since np < 5 and nq < 5. What is the value of the sample test statistic? (Round your answer to two decimal places.) (iii) Find (or estimate) the P-value. P-value > 0.500 0.250 < P-value < 0.500 0.100 < P-value < 0.250 0.050 < P-value < 0.100 0.010 < P-value < 0.050 P-value < 0.010 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 in parts (i) to (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 statistically significant. 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 statistically significant. At the α = 0.05 level, we fail to reject the null hypothesis and conclude the data are not statistically significant. (v) Interpret your conclusion in the context of the application. There is sufficient evidence at the 0.05 level to conclude that there is a difference from the national average for the population proportion of crime victims. There is insufficient evidence at the 0.05 level to conclude that there is a difference from the national average for the population proportion of crime victims. (b) Find a 95% confidence interval for the proportion of students in this school (ages 16-19) who have been victims of a crime. lower limit upper limit (c) How large a sample size should be used to be 95% sure that the sample proportion p̂ is within a margin of error E = 0.04 of the population proportion of all students in this school (ages 16-19) who have been victims of a crime? Hint: Use sample data p̂ as a preliminary estimate for p. (Round your answer up to the nearest student.) students

Answer 1

(a)

Here we have x=9, n=66,p 0.1364 7 Hypotheses are Ho : p = 0.1 Ha.p#0.1 Level of significance : =0.05 α Test is two tailed The standard deviation is: sqrt((0.1*(1-0.1))/66) = 0.0369 Test Statistics p-p =(0.1364-0.1)/(0.0369) -0.99 Critical value: CV = 1.96 Rejection Region: If |z| > 1.96, Reject HO Decision: Since test statistics does not lie in rejection region so we fail to reject the null hypothesis P-value-0.3222 P-value: Decision: Since p-value is not less than level of significance so we fail to reject the null hypothesis Excel function for critical value: NORMSINV(1-(0.05/2)) 2*(1-NORMSDIST (0.99)) Excel function for p-value:

(a) (i) What is the level of significance? 0.05

The null and alternate hypotheses:

H0: p = 0.10; H1: p ≠ 0.10

(ii) The standard normal, since np > 5 and nq > 5.

The test statistics is: z =0.99

(iii) The p-value is 0.3222

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

(v)There is insufficient evidence at the 0.05 level to conclude that there is a difference from the national average for the population proportion of crime victims.

(b)

Here we have -9, n=66, p= :0.1364 n x Level of significance: Critical value: α-005 1.96 The standard deviation is: Pli-p) =sqrt((0.1364*(1-0.1364)/66) 0.0422 The requried confidence interval is: ptz, o 0.1364 (1.96 0.0422) 0.1364 t 0.0827 (0.054,0.219) Excel function for critical value: NORMSINV(1-(0.05/2)) Hence, the required confidence interval is (0.054,0.219)

(c)

Here we have ME-0.04,p -0.1364, CL-0.95 Level of significance: -0.0s Critical value of z is 1.96 "I-a /2 Required sample size is: 1-a/2 -((1.96A2*0.1364 (1-0.1364)/0.0442 7 ME = 282.82589104 Hence, the required sample size is 283. Excel function for critical value: NORMSINV(1-(0.2/2))