Question

1. A normal population has a mean of µ = 100. A sample of n = 36 is selected from the population, and a treatment is administered to the sample. After treatment, the sample mean is computed to be M = 106. Assuming that the population standard deviation is σ = 12, use the data to test whether or not the treatment has a significant effect. Use a one tailed test.

Hypothesis:                                        Z_crit =

z test calculation:                                                                 

Conclusion:

Answer 1

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Answer:

The null and alternative hypothesis for this test is given as below:

H₀: µ = 100 versus H_a: µ ≠ 100

Level of significance = alpha = 0.05

Test statistic formula is given as below:

Z = (Xbar - µ) / (σ/sqrt(n)]

Z = (106 – 100) / [12/sqrt(36)]

Z = 6/[12/6] = 6/2 = 3

zcrit = 1.96

P-value = 0.0027

Alpha value = 0.05

z test conclusion - Observed z value = 3 > zcrit of 1.96 Hence we reject the null hypothesis

We conclude that there is sufficient evidence that the treatment has a significant effect.