Hypothesis Testing: One Sample Mean t

Introduction

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

• You are dealing with a single sample mean (\(\bar{x}\)).
• The SAMPLE standard deviation (\(s\)) is known, but the population standard deviation is NOT

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. 
• \(s\) refers to the sample standard deviation.
• \(n\) refers to the sample size.

Example

At the Chocolate Factory, an overseer supervises thousands of employees. He claims that his employees each make 130 candy bars on average per hour. The Chocolate Factory boss disagrees with this overseer; he thinks that the employees produce less than 130 candy bars per hour. He sets out to prove this by collecting a random sample of 46 employees and recording how many candy bars each produces in an hour. From this data, he calculates a sample mean of 112 and sample standard deviation of 17. Conduct a hypothesis test at the significance level of 0.05 to determine whether the Chocolate Factory workers are actually producing 130 candy bars per hour. 

Step 1: Name Test: 1 Sample Mean t 

Step 2: Define Test: 

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

In this case, we believe that the average candy bar production is less than what the overseer has stated, so we are conducting a left sided test:

\(H_0: \mu = 130\)

\(H_A : \mu < 130\)

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:

\(t = {112 -130 \over {17 \over \sqrt {46} } }\)  →  \(t = -7.18\)

P-Value: 

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

• lower limit: -999
• upper limit: \(t\)
• degrees of freedom: \(n-1\)
• All together, it looks like this: tcdf (-999, \(t\), \(n-1\))

*Note: If it was a right-sided test and the test statistic was positive (t > 0), then your lower limit would be the test statistic (t) and your upper limit would be 999. 

In this case, we do tcdf (-999, -7.18, 45) to get a p-value of 2.76 \(\times\) 10^-9 (approximately zero).

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

We calculated a p-value of approximately zero. This p-value is less than the assumed significance level of 0.05. Therefore, we reject the null hypothesis. The data supports the claim that the average number of candy bars produced per hour by employees at the Chocolate Factory is less than 130.