Question

A sample of 44 observations is selected from a normal population. The sample mean is 46, and the population standard deviation is 7. Conduct the following test of hypothesis using the 0.01 significance level.

H₀: μ = 50

H₁: μ ≠ 50

1. Is this a one- or two-tailed test?

• One-tailed test

• Two-tailed test

2. What is the decision rule?

• Reject H₀ if −2.576 < z < 2.576

• Reject H₀ if z < −2.576 or z > 2.576

3. What is the value of the test statistic? (Negative amount should be indicated by a minus sign. Round your answer to 2 decimal places.)

4. What is your decision regarding H₀?

• Fail to reject H₀

• Reject H₀

1. e-1. What is the p-value? (Round your z value to 2 decimal places and final answer to 4 decimal places.)

1. e-2. Interpret the p-value? (Round your z value to 2 decimal places and final answer to 2 decimal places.)

There is a _____ % chance of finding a z value this large by "sampling error" when H0 is true.

rev: 01_21_2018_QC_CS-114345

Answer 1

Answer:

a) Two-tailed test

b) Reject H0 if z < −2.576 or z > 2.576

c) test statistic =   Z = -3.79

d) Reject H0

e_1) p-value = 0.0000

e_2)   There is a 0 % chance of finding a z value this large by "sampling error" when H0 is true.

Given information is n = sample size = 44

= sample mean =46

= population standard deviation = 7

= significance level = 0.01

a) Our alternative hypothesis is  H1: μ ≠ 50 . this is two tailed

one tailed hypothesis is either or

b) we have = significance level = 0.01,   = 2.756

And we reject H0 if Z > 2.756 or Z < -2.756

c)   test statistic is

  

Z = -3.79 ( rounded to 2 decimals )

d) From the decision criteria we have Z = -3.79 < -2.756 , hence we reject H0

e_1 ) To find p- value use excel function : =NORMSDIST(-3.79) = 0.0000753

= 0.0000 ( rounded to 4 decimals)

e_2) As our p-value is 0 hence we interpret it as,

There is a 0 % chance of finding a z value this large by "sampling error" when H0 is true.

MINITAB output of one sample z test for reference

One-Sample Z

Test of μ = 50 vs ≠ 50
The assumed standard deviation = 7

N Mean SE Mean 99% CI Z P
44 46.00 1.06 (43.28, 48.72) -3.79 0.000
(Z and P-value in bold)