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The z-test is the ideal hypothesis test to conduct in a **normal distribution** of a random variable. In addition, the variance of the population must be **known**. The z-statistic refers to the test statistic computed for hypothesis testing.

**Testing H _{0}: μ = μ_{0} using the z-test**

Given a random sample of size *n* from a normally distributed population with mean μ and variance σ^{2, }and a sample mean X̄, we can compute the z-statistic as:

$$ \text z-\text{statistic} =\cfrac {(\bar{X} – \mu_0)}{\left(\frac {\sigma}{\sqrt n} \right)} $$

Where:

\(\bar{X}\) is the sample mean.

\(μ_0\) is the hypothesized mean of the population.

\(σ\) is the standard deviation of the population.

\(n\) is the sample size.

Once computed, the z-statistic is compared to the critical value that corresponds to the level of significance of the test. For example, if the significance level is 5%, the z-statistic is screened against the upper or lower 95% point of the normal distribution (±1.96). The decision rule is to reject the H_{0 }if the z-statistic falls within the critical or rejection region.

Academics carried out a study on 50 former United States presidents and found an average IQ of 135. You are required to carry out a 5% statistical test to determine whether the average IQ of presidents is greater than 130. (IQs are distributed normally, and previous studies indicate that σ = 25.)

**Solution**

First, you have to state the hypothesis:

H_{0}: μ = 130

H_{1}: μ > 130

Assuming the H_{0} is true, \(\frac {(\bar{X} – 130)}{\left(\frac {\sigma}{\sqrt n} \right)} \sim N(0,1)\).

The z-statistic is \(\cfrac {(135 – 130)}{\left(\frac {25}{\sqrt {50}} \right)} = 1.414\).

This is a right-tailed test. Therefore, we compare our test statistic to the upper 95% point of the standard normal distribution (1.6449). Since 1.414 is less than 1.6449, we **do not have** sufficient evidence to reject the H_{0}. As such, it would be **reasonable **to conclude that the average IQ of U.S. presidents is not more than 130.

The t-test is based on the t-distribution. The test is appropriate for testing the value of a population mean when:

- σ is unknown; and
- The sample size is large (n ≥ 30), and if n < 30, the distribution must either be normal or approximately normal.

We compute a t-statistic with n – 1 degrees of freedom as:

$$ t_{n-1} = \cfrac {(\bar{X} – \mu_0)}{\left(\frac {s}{\sqrt n} \right)} $$

Where:

\(\bar{X}\) is the sample mean.

\(μ_0\) is the hypothesized mean of the population.

\(s\) is the standard deviation of the sample.

\(n\) is the sample size.

The annual rate of rainfall (cm) in a certain equatorial country over the last 10 years is given below:

{ 25 26 25 27 28 29 28 27 26 25 }

Financial analysts in the country wish to determine whether the average rate of rainfall has increased from its former value of 23. Carry out a statistical test at the 5% level.

**Solution**

As always, you should begin by stating the hypothesis:

H_{0}: μ ≤ 23

H_{1}: μ > 23

If we assume that the annual rainfall quantities are distributed normally and recorded independently, then:

$$ \cfrac {(\bar{X} – 23)}{\left(\frac {S}{\sqrt n} \right)} \sim t_{n- 1} $$

Please, confirm that X̄ = 26.6 and S = 1.43.

Therefore, our t-statistic = \(\cfrac {(26.6 – 23)}{\left(\frac {1.43}{\sqrt {10}}\right)} = 7.96\).

Our test statistic (7.96) is greater than the upper 95% point of the t_{0.05,9} distribution (1.833).

Therefore, we have **sufficient evidence** to reject the H_{0}. As such, it is **reasonable** to conclude that the average annual rainfall has increased from its former long-term average of 23.

QuestionWhat is the value of

tin the example above if the level of significance is reduced from 5% to 0.5%, and does this change the decision rule?

- 2.02; it does not change the decision rule.
- 3.25; it does not change the decision rule.
- 3.25; it does change the decision rule.

SolutionThe correct answer is

B.A quick glance at the t

_{0.005,9}distribution when α = 0.5% gives a value of 3.25.However, the evidence against the H

_{0}is too strong since our test statistic (7.96) is still greater than 3.25. As such, the conclusion would remain unchanged.

*(***Note to candidates:*** You might also work out the solutions to the above examples and questions using p-values instead of critical values. The decision rules would remain unchanged provided you work out the p-values correctly.)*