Question

9. In a test of hypotheses of the form ?0 : μ = 78.3 versus H1 : μ ≠78.3 using ? = .01, when the sample size is 28 and the population is normal but with unknown standard deviation the rejection region will be the interval or union of intervals: (a) [2.771, ∞) (b) [2.473, ∞) (c) [2.576, ∞) (d) (−∞,−2.771] ∪ [2.771, ∞) (e) (−∞,−2.473] ∪ [2.473, ∞)

10. In a test of hypotheses ?0: μ = 2380 versus ?1 : μ > 2380, the rejection region is the interval [2.718, ∞), the value of the sample mean computed from a sample of size 12 is ?̅= 2413, and the value of the test statistic is t = 2.902. The correct decision and justification are: (a) Do not reject H0 because the sample is small. (b) Do not reject H0 because 2.718 < 2.902. (c) Reject H0 because 2413 is larger than 2380. (d) Reject H0 because 2.902 is positive. (e) Reject H0 because 2.902 lies in the rejection region.

11. In a test of hypotheses of the form ?0 : p = .62 vs. ?1 : p < .62 a sample of size 1200 produced the test statistic z = −1.915. The p-value (observed significance) of the test is about: (a) 0.06 (b) 0.03 (c) 0.97 (d) −0.03 (e) 0.57

Answer 1

9)

?0 : μ = 78.3

H1 : μ ≠78.3

n=28, df= n-1 = 28-1 =27

using t table we get rejection region as follows

(−∞,−2.771] ∪ [2.771, ∞)

10)

?0: μ = 2380

?1 : μ > 2380

sample size (n) = 12

the rejection region is the interval [2.718, ∞)

test statistic is t = 2.902

since ( test statistic= 2.902) > ( critical value = 2.718)

Hence,

(e) Reject H0 because 2.902 lies in the rejection region.

11)

?0 : p = .62 vs. ?1 : p < .62

sample size (n) =1200

test statistic z = −1.915

P-Value = P(z < -1.92)

using normal table we get

P(z < -1.92) = 0.0274

P-Value = 0.03