Question

Consider the following hypotheses:

H0: μ ≤ 270

HA: μ > 270

Find the p-value for this test based on the following sample information. (You may find it useful to reference the appropriate table: z table or t table)

a. x¯x¯ = 277; s = 23; n = 18

  • 0.025   p-value < 0.05
  • 0.01   p-value < 0.025
  • p-value    0.10
  • 0.05   p-value < 0.10
  • p-value < 0.01

b. x¯x¯ = 277; s = 23; n = 36

  • p-value   0.10
  • 0.025   p-value < 0.05
  • 0.05   p-value < 0.10
  • p-value < 0.01

  • 0.01   p-value < 0.025

c. x¯x¯ = 277; s = 14; n = 11

  • 0.05   p-value < 0.10
  • p-value   0.10
  • p-value < 0.01

  • 0.01   p-value < 0.025

  • 0.025   p-value < 0.05

d. x¯x¯ = 275; s = 14; n = 11

  • 0.01   p-value < 0.025
  • p-value < 0.01

  • p-value   0.10
  • 0.05    p-value < 0.10
  • 0.025   p-value < 0.05
0 0
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Answer #1


Date: 11/11/2019 Answer The null and alternative hypothesis is, Ho u 270 Hu270 a a) The f-test statistics is, t= /Jn 277 270d.f. n-1 =18 -1 17 and f-test statistics is 2.284 and the p-value is between 0.01 and 0.025. The p-value for this test is 0.0From the t distribution table, show the degree of freedom is, d.f. n- - 36 -1 35 and f-test statistics is 3.231 and the p-valThe p-value for this test is, From the f distribution table, show the degree of freedom is, d.f.- n-1 11-1 10 and f-test statThe f-test statistics is 1.185 The p-value for this test is, From the t distribution table, show the degree of freedom is, d.

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