# Hypothesis Testing

To perform a hypothesis test, there are four steps that must be followed.

1. State the null **H _{0}** and the alternative

**H**hypothesis.

_{1}Two Tailed | H_{0}= k |
H_{1} ≠ k |
---|---|---|

One Tailed | H_{0 }≥ k |
H_{1} < k |

H_{0} ≤ k |
H_{1} > k |

2.From a confidence level, **(1 − α)**, or a significance level, **α**, determine:

The value of **z _{α/2}** (two tailed), or

**z**(one tailed)

_{α} The limit of acceptance of the sample parameter (**x o p'**).

3.Calculate: the value of** μ **or **p** from the sample.

4.If the sample parameter value is within the limit of acceptance, accept the hypothesis with a significance level **α**.

If not, reject it.

## Two Tailed Test

A two-tailed test occurs when the null hypothesis is of the type **H _{0}: μ = k** (or

**H**) and the alternative hypothesis, therefore, is of the type

_{0}: p = k**H**(or

_{1}: μ≠ k**H**).

_{1}: p≠ k The significance level, **α**, is concentrated in two parts (or tails) symmetrical about the mean.

The limit of acceptance in this case is the corresponding confidence interval for μ or p, that is to say:

or:

#### Example

It is known that the standard deviation of the scores in a math exam was 2.4 and a sample of 36 students scored an average of 5.6. With this data, can the hypothesis be confirmed that the average test score was 6 with a confidence level of 95%?

1. State the null and alternative hypotheses:

**H _{0} : μ = 6 ** The average test score has not varied.

**H _{1} : μ ≠ 6 ** The average test score has varied.

2. Calculate the limit of acceptance:

For a significance level of **α = 0.05**, the corresponding critcal value is: z_{α/2} = 1.96.

Calculate the confidence interval for the mean:

** (6 − 1.96 · 0.4, 6 + 1.96 · 0.4) = (5.22, 6.78)**

3. Verify:

The value of the mean of the sample is:** 5,6 **.

4. Decide:

The nule hypothesis, H_{0}, should be accepted with a confidence level of 95%.

## One Tailed Test

#### Case 1

The nule hypothesis is of the type** H _{0}: μ ≥ k ** (or

**H**).

_{0}: p ≥ kThe alternative hypothesis, therefore, is of the type **H _{1}: μ < k** (or

**H**).

_{1}: p < k### Critical Values

1 - α | α | z_{α} |
---|---|---|

0.90 | 0.10 | 1.28 |

0.95 | 0.05 | 1.645 |

0.99 | 0.01 | 2.33 |

The significance level, **α**, is concentrated in one part or tail.

The limit of acceptance in this case is:

or:

#### Example

A sociologist has predicted that in a given city, the level of absenteeism in the upcoming elections will be a minimum of 40%. From a random sample of 200 individuals from the voting population, 75 state they will likely vote. Determine with a significance level of 1%, if the hypothesis can be accepted.

1. State the null and alternative hypotheses:

**H _{0} : p ≥ 0.40 ** The absenteeism will be a minimum of 40%.

**H _{1} : p < 0.40** The absenteeism will be a maximum of 40%.

2. Calculate the limit of acceptance:

For a significance level of **α = 0.01**, the corresponding critcal value is: z_{α} = 2.33.

Determine the confidence interval:

3. Verify:

4. Decide:

The nule hypothesis, H_{0}, should be accepted as it can be stated with a significance level of 1% that absenteeism will be at least 40% for the upcoming election.

#### Case 2

The nule hypothesis is of type **H _{0}: μ ≤ k** (or

**H**).

_{0}: p ≤ kThe alternative hypothesis is, therefore, of type **H _{1}: μ > k** (or

**H**).

_{1}: p > kThe significance level, **α**, is concentrated in one part or tail.

The limit of acceptance in this case is:

or:

#### Example

A report indicates that the maximum price of a plane ticket between New York and Chicago is $120 with a standard deviation of $40. A sample of 100 passengers shows that the average price of their tickets was $128.

Can the above statement be accepted with a significance level equal to 0.1?

1. State the null and alternative hypotheses:

**H _{0} : μ ≤ 120 **

**H _{1} : μ > 120 **

2. Calculate the limit of acceptance:

For a significance level of **α = 0.1**, the corresponding critical value is: z_{α} = 1.28.

Calculate the confidence interval for the mean:

3. Verify:

The value of the mean of the sample is:** $128**.

4. Decide:

**The nule hypothesis, H _{0},** cannot be accepted with a significance level equal to 0.1.