A test of hypotheis entails:

- Calculating the z-score.
- (Comparing the z-score to the critical value, which we will not do, or) coverting the z-score to a p-value.
- Comparing the p-value to the significance level.

- The z-score is ((x-bar) - µ)/(*sigma*/(n^.5)); the numerator is the difference between the observed and hypothesized mean, the denominator rescales the unit of measurement to standard deviation units. (95-85)/(20/(22^.5)) = 2.3452.
- The z-score 2.35 corresponds to the probability .9906, which leaves
.0094 in the tail beyond. Since one could have been as far below 85, the
probability of such a large or larger z-score is .0188. This is the p-value.
Note that for these two tailed tests we are using the absolute value of the
z-score.

- Because .0188 < .05, we reject the hypothesis (which we shall call the null hypothesis) at the 5% significance level; if the null hypothesis were true, we would get such a large z-score less than 5% of the time. Because .0188 > .01, we fail to reject the null hypothesis at the 1% level; if the null hypothesis were true, we would get such a large z-score more than 1% of the time.

- You reject the null hypothesis if the z-score is large, which means that the p-value is small.
- If you reject a hypothesis at the 5% significance level, p < .05, hence you will reject that hypothesis at the 10% significance level.
- If you fail to reject a hypothesis at the 5% significance level, p
> .05, hence you will fail to reject that hypothesis at the 1%
significance level.
**Competencies:**If the standard deviation is known to be equal to 12, and your null hypothesis is that the mean of the population is equal to 15, at what level (p-value) is x-bar = 13.5 based on a sample of size 200 significant? Would you reject the null hypothesis at the 10% significance level? 5% significance level? 1% significance level?**Reflection:**What is the relationship between a two tailed test of hypothesis and a confidence interval?