Question

Assume that the differences are normally distributed. Complete parts (a) through (d) below. 2 3 4 5 6 7 8 Observation X; 54.0 46.0 43.9 40.3 48.3 44.4 48.4 42.7 54.6 46.8 46.5 45.5 50.5 46.1 50.7 45.6

(a) Compute avg(d) and s_d.

(b) Test if μ_d<0 at the α = 0.01 level of significance. Round to five decimal places as needed.

(c) Test if μ_d<0 at the α = 0.05 level of significance. Round to five decimal places as needed.

(d) Compute a​ 90% confidence interval about the population mean difference μ_d. Round to five decimal places as needed.

(e) Compute a​ 95% confidence interval about the population mean difference μ_d. Round to five decimal places as needed.

(f) Compute a​ 98% confidence interval about the population mean difference μ_d. Round to five decimal places as needed.

Answer 1

Let us denote the difference

d = X- Y

a)

b)

Let ud denotes the population mean of difference. Hypotheses : To test null hypothesis Ho Jd = 0 against alternative hypothesis H1 :Hd < 0 This is a one sided left tailed test as alternative hypothesis H1 contains only < sign. Let à denotes the sample mean of difference, sd denotes the sample standard deviation of difference and n denotes the sample size. Here, d = -2.2875, să = 2.0527, sa =1.4327, n=8 The test statistic can be written as: Vnd – 0.0) t- sd which under H, follows a t distribution with n - 1 df. Decision rule / Rejection region : We reject H, at 0.01 level of significance if P-value < 0.01 or if tobs <-+0.01,1–1 =-2.997952 0.4- 0.3 0.2- 0.1 0.0 0 t Now, Value of test statistic: The value of the test statistic tobs 18(-2.2875 - 0) 1.4327 = -4.515914 -4.52 associated degrees of freedom = n -1 = 7 The critical value =tcritical = -0.01,7 = -2.997952 -2.998 P-value = P(tn-1 <tobs) = P(t7 -4.515914) = 0.001373 0.0014 Conclusion : Since p-value < 0.01 and tobs <tcritical = -2.997952, so we reject H, at 0.01 level of significance Hence, we can conclude that mean difference is significantly less than 0.

Let ud denotes the population mean of difference. Hypotheses : To test null hypothesis Ho ipd = 0 against alternative hypothesis Hj Hd < 0 This is a one sided left tailed test as alternative hypothesis H¡ contains only < sign. Let d denotes the sample mean of difference, sa denotes the sample standard deviation of difference and n denotes the sample size. Here, d = -2.2875, să = 2.0527, sa = 1.4327, n = 8 The test statistic can be written as: vnd – 0.0) t= sd which under Ho follows at distribution with n-1 df. Decision rule / Rejection region : We reject H, at 0.05 level of significance if P-value < 0.05 or if tobs < -+0.05,n-1 =-1.894579 0.4- 0.3 20.2 0 0.1 0.0 -2 t Now, Value of test statistic: The value of the test statistic tobs 18(–2.2875 - 0) 1.4327 = -4.515914 -4.52 associated degrees of freedom = n -1 = 7 The critical value =tcritical = -10.05,7 = -1.894579 -1.895 P-value = P(tn-1 < tobs) = P(tz < -4.515914) = 0.001373 -0.0014 Conclusion : Since p-value < 0.05 and tobs <tcritical = -1.894579, so we reject H, at 0.05 level of significance Hence, we can conclude that mean difference is significantly less than 0.

c)

Let d denotes the sample mean of difference, s denotes the sample standard deviation of difference and n denotes the sample size. Here, Sd 1.4327 d = -2.288, să 2.0527, sd = 1.4327, n = 8, standard error of mean 0.506542 In 78 The margin of error is Sd 1.4327 1.4327 ME *t 1-0.9 * 10.05,7 * 1.894579 = 0.959684 ~ 0.960 n 78 a 90 percent confidence interval for the population mean of difference ud is Sd = d = *t 1-0.9 ✓n ,7 2 1.433 = -2.288 + * 10.05,7 18 = -2.288 + 1.433 * 1.894579 18 = -2.288 + 0.959684 = (-3.247184, -1.327816) = (-3.24718, -1.32782)

d)

e)

Here, sd 1.4327 d = -2.288, số = 2.0527, sd = 1.4327, n = 8, standard error of mean 0.506542 ✓n 18 The margin of error is Sd ME *t 1-0.95 1.4327 V8 * 10.025,7 1.4327 *2.364624 = 1.197782 = 1.198 18 vn ,7 2 a 95 percent confidence interval for the population mean of difference ud is Sd =d + = * t 1-0,95,7 vñ 1.433 = -2.288 + * 10.025,7 18 1.433 = -2.288 + * 2.364624 18 = -2.288 +1.197782 = (-3.485282, -1.089718) z (-3.48528, -1.08972)

Here, Sd 1.4327 d = -2.288, să = 2.0527, sa = 1.4327, n = 8, standard error of mean = 0.506542 ✓n The margin of error is sd 1.4327 ME *t 1-0.98 *t0.01,7 1.4327 * 2.997952 = 1.518588 ~ 1.519 18 V8 a 98 percent confidence interval for the population mean of difference ud is Ed + Sd = * † 1-0,98,7 vn 1.433 = -2.288 + * to.01,7 1.433 = -2.288 + * 2.997952 18 = -2.288 + 1.518588 = (-3.806088,-0.768912) ~(-3.80609, -0.76891)