Hypothesis Testing: One Sample Mean Z

Introduction

One Sample Mean Z hypothesis tests are used when...

• You are dealing with a single sample mean ($$\bar{x}$$)
• The POPULATION standard deviation ($$\sigma$$) is known

A few symbols need to be defined before we dive in:

• $$\mu$$ refers to the given mean that is assumed to be true.
• $$\bar{x}$$ refers to the sample mean that you will use to disprove the null.
• $$\sigma$$ refers to the population standard deviation.
• $$n$$ refers to the sample size.

Example

At Latin High School, the guidance department claims that the mean GPA is 3.2, but you feel that the mean is greater than 3.2. You set out to prove this by collecting a random sample of 36 students, which produces a sample mean of 3.4. It is known that the standard deviation of the population is 0.4. Conduct a hypothesis test at the significance level of 0.05 to determine whether the Latin High School GPA is actually 3.2

Step 1: Name Test: 1 Sample Mean Z

Step 2Define Test:

For a one sample mean hypothesis test where the standard deviation of the population ($$\sigma$$) is known, the following three alternative test options are available:

 Left-Sided Test Two-Sided Test Right-Sided Test $$H_{o}\!\!: \mu = \mu_{o}$$ $$H_{a}\!\!: \mu < \mu_{o}$$ $$H_{o}\!\!: \mu = \mu_{o}$$ $$H_{a}\!\!: \mu \neq \mu_{o}$$ $$H_{o}\!\!: \mu = \mu_{o}$$ $$H_{a}\!\!: \mu > \mu_{o}$$

In this case, we believe that the true GPA is greater than what the high school has stated, so we are conducting a right sided test:

$$H_0 : \mu = 3.2$$

$$H_A : \mu > 3.2$$

Step 3: Assume $$H_0$$ is true and define its normal distribution. Then check the conditions.

1. The data is a random sample for the population.

2. N ≥ 10n

3. The sampling distribution of $$\bar{x}$$ is approximately normal.

Note: This can be easily checked using the Central Limit Theorem.

Step 4: Using the normal distribution, calculate the test statistics and p-value.

Test Statistic (1 Sample Mean z):  $$z = \frac{\bar{x}\, -\, \mu_{o}}{\sigma \, /\sqrt{n}}$$

Test Statistic:

$$z = {3.4 - 3.2 \over {0.4 \over \sqrt {36}}}$$  →  $$z = 3.0$$

P-Value:

The p-value will be found by using the normal cdf function on your calculator:

• lower limit: $$z$$
• upper limit: 999
• distribution center: 0
• standard deviation: 1
• All together, it looks like this: normalcdf ($$z$$, 999, 0, 1)

*Note: If it was a left-sided test and the test statistic was negative (z < 0), then your lower limit would be -999 and your upper limit would be the test statistic (z).

In this case, we do normalcdf (3, 999, 0, 1) to get a p-value of 0.001

Step 5: Analyze your results and determine if they are statistically significant.

We calculated a p-value of 0.001. This p-value is less than the significance level of 0.05. Therefore, we reject the null hypothesis. The data supports the claim that the mean GPA at Latin High School is greater than 3.2.