Question

Consider the following hypotheses: H0: μ = 61 HA: μ ≠ 61 Approximate the p-value for this test based on the following sample information. Use Table 2. a. x¯ = 58; s = 10.3; n = 10 0.20 < p-value < 0.40 0.10 < p-value < 0.20 0.05 < p-value < 0.10 p-value < 0.05 p-value Picture 0.4 b. x¯ = 64; s = 10.3; n = 10 0.20 < p-value < 0.40 0.10< p-value < 0.20 0.05 < p-value < 0.10 p-value < 0.05 p-value Picture 0.4 c. x¯ = 57; s = 8.9; n = 13 0.10 < p-value < 0.20 0.01 < p-value < 0.03 0.05 < p-value < 0.10 p-value < 0.01 p-value Picture 0.2 d. x¯ = 58; s = 8.9; n = 19 0.10 < p-value < 0.20 0.01 < p-value < 0.03 0.05 < p-value < 0.10 p-value < 0.01 p-value Picture 0.2

Answer 1

Solution :

This is the two tailed test .

The null and alternative hypothesis is ,

H₀ :   = 61

H_a :    61

(a)

= 58

= 61

s = 10.3

n = 10

Degrees of freedom = n - 1 = 10 - 1 = 9

Test statistic = t

= ( - ) / s / n

= (58 - 61) / 10.3 / 10

= -0.92

Test statistic = -0.92

0.20 < p-value < 0.40

(b)

= 64

= 61

s = 10.3

n = 10

Degrees of freedom = n - 1 = 10 - 1 = 9

Test statistic = t

= ( - ) / s / n

= (64- 61) / 10.3 / 10

= 0.92

Test statistic = 0.92

0.20 < p-value < 0.40

(c)

= 57

= 61

s = 8.9

n = 13

Degrees of freedom = n - 1 = 13 - 1 = 12

Test statistic = t

= ( - ) / s / n

= (57 - 61) / 10.3 / 10

= -1.62

Test statistic = -1.62

0.10 < p-value < 0.20

(d)

= 58

= 61

s = 8.9

n = 19

Degrees of freedom = n - 1 = 19 - 1 = 18

Test statistic = t

= ( - ) / s / n

= (58 - 61) / 10.3 / 10

= -1.47

Test statistic = 1.47

0.10 < p-value < 0.20