T-Formula & Null Hypotheses

Homework #3

9.11

A random sample of size 20 is taken, resulting in a sample mean of 16.45 and a sample standard deviation of 3.59. Assume x is normally distributed and use this information and α = .05 to test the following hypotheses.

 

            H0: µ = 16

            Ha: µ ≠ 16

 

Since both of the following statements is not true with respect to this problem:

 

  1. x is normally distributed
  2. n > 30

 

We do not use the z-formula; we use the t-formula:

 

 

 

df = n - 1

 

Decision Rule:

 

The test is two tailed as Ha uses the not equal assignment.

Confidence level is 1 – α = 1 – .05 = .95

 

As this is a two-tailed test, there is α/2 or .025 area on each end of the distribution which is the rejection region. Another way to put it is the rejection region covers 2.5% (.025) of each of the two ends of the distribution (total 5%) leaving 95% (.4750) in the non-rejection region.

 

There is a .4750 area between the mean and each of the critical values that separate the rejection region from the non-rejection region.

 

From Table A.5 and the .4750 area, the critical z value zα/2 is ±1.96.

 

Decision Rule:

If the data gathered produces a z value greater than +1.96 or less than -1.96, the decision is to NOT reject the null hypothesis because the observed z value is in the non-rejection region.

 

Calculate the z value, given:

 

= 16.45

 

 

 n = 20

df = n – 1 = 19

 

 

 

tobserved = .56

 

For two-tailed tests α/2 = .025, therefore from t-Table Tα/2,df   = T.025,19 = ±2.093.

 

Now, because tobserved (.56) is less than tTable (2.093), we do NOT reject the null hypotheses H0.