Question

A. Verify that the method is appropriate, write out formula for your t-statistic, and explain/show the steps you use to get your value of t. (If you solved this with JMP or Excel Data Toolpak, that is fine, but please either explain what you did or provide screenshots of your input)

B. Explain (possibly with a sentence) how you obtained your p-value, and then explain how your p-value led you to your final conclusion. (Even if you used technology, explain how the p-value could be obtained from the t-statistic and how that p-value leads you to your conclusion

Answer 1

I'm taking a example of one sample t test

The one sample t test compares the mean of your sample data to a known value. For example, you might want to know how your sample mean compares to the population mean. You should run a one sample t test when you don’t know the population standard deviation or you have a small sample size.

Assumptions of the test (your data should meet these requirements for the test to be valid):

• Data is independent.
• Data is collected randomly.
• The data is approximately normally distributed.

One Sample T Test Example

Your company wants to improve sales. Past sales data indicate that the average sale was $100 per transaction. After training your sales force, recent sales data (taken from a sample of 25 salesmen) indicates an average sale of $130, with a standard deviation of $15. Did the training work? Test your hypothesis at a 5% alpha level.

Step 1: Write your null hypothesis statement (How to state a null hypothesis). The accepted hypothesis is that there is no difference in sales, so:
H0: μ = $100.

Step 2: Write your alternate hypothesis. This is the one you’re testing. You think that there is a difference (that the mean sales increased), so:
H1: μ > $100.

Step 3: Identify the following pieces of information you’ll need to calculate the test statistic. The question should give you these items:

1. The sample mean(x̄). This is given in the question as $130.
2. The population mean(μ). Given as $100 (from past data).
3. The sample standard deviation(s) = $15.
4. Number of observations(n) = 25.

Step 4: Insert the items from above into the t score formula.



t = (130 – 100) / ((15 / √(25))
t = (30 / 3) = 10
This is your calculated t-value.

Step 5: Find the t-table value. You need two values to find this:

1. The alpha level: given as 5% in the question.
2. The degrees of freedom, which is the number of items in the sample (n) minus 1: 25 – 1 = 24.

Look up 24 degrees of freedom in the left column and 0.05 in the top row. The intersection is 1.711. This is your one-tailed critical t-value.

What this critical value means is that we would expect most values to fall under 1.711. If our calculated t-value (from Step 4) falls within this range, the null hypothesis is likely true.

Step 5: Compare Step 4 to Step 5. The value from Step 4 does not fall into the range calculated in Step 5, so we can reject the null hypothesis. The value of 10 falls into the rejection region (the left tail).

-------P value Method

P value = P(t(24) > 10)

Use, the t table at 24 degree of freedom, p value = 1- P(t(24) < 10) = 1-0.999999999 = almost 0

Since, p value is less than alpha, so we reject null hypothesis

In other words, it’s highly likely that the mean sale is greater. The sales training was probably a success.