Question

- 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:**

**H _{0}:
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.**

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