
When performing a hypothesis test, we are using sampled data to draw a conclusion about the population. It’s possible that our conclusion will be incorrect.

What is the probability of making a Type 1 Error? \(\alpha\), the predetermined significance level
Determining the probability of making a Type 2 Error is not as straight forward. We won’t focus on this. We will focus on…
the fact that Type 1 and Type 2 errors are related.
Please locate the Class 14 Activity in Top Hat and answer Questions 1 and 2.
Motivating Example: A study is done by a community group to determine if the students at two different universities (A and B) graduate having taken a different number of math courses, on average.
The data: To answer this question, the group randomly samples students from each university.
University A samples 46 graduates. Their average is four math classes with a standard deviation of 1.5 math classes.
University B samples 32 graduates. Their average is 3.5 math classes with a standard deviation of one math class.
Calculate the point estimate, \(\overline{x}_1 - \overline{x}_2\)
Identify the confidence level, \(CL\), and the error associated with this confidence level \(100-CL\).
Determine the critical value, \(t^*\), by finding the \(CL + \frac{100-CL}{2}\) percentile on the t distribution with \(\nu\) degrees of freedom (df).
Calculate the standard error estimate from the observed sample:
\[\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}\]
Link to this calculator and an R script alternative is available in the Week 7 Canvas module.
qt()
qt(p, df) calculates the value on a t distribution curve with df degrees of freedom that has an area of p to the left of it.
Example: To find the critical value needed to construct a 99% confidence interval for the difference in means when \(s_1 = 1.5\), \(n_1 = 46\), \(s_2 = 1\), \(n_2 = 32\):
qt(0.995, 75.897) = 2.642
(\(\overline{x}_1 - \overline{x}_2)\) \(\pm\) \(t^*_{\nu}\) \(\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}\)
Motivating Example:
(\(4-3.5)\) \(\pm\) \(2.645\) \(\sqrt{\frac{1.5^2}{46}+\frac{1^2}{32}}\)\(= (-0.248, 1.248)\)
We can perform a formal hypothesis test to answer questions concerning a difference in two population means.
We’ll use the same steps introduced in our Introduction to Hypothesis Testing notes, but some of the details will differ in this new scenario.
Motivating Example:
Is there a difference in the average number of math courses taken between graduates at University A and University B?
Motivating Example:
The difference in the average number of math courses taken, \(\mu_A - \mu_B\)
Null hypothesis
\[H_0: \mu_1 = \mu_2\]
Alternative hypothesis:
This depends on the question of interest.
Lower one-sided
Question of interest: Is the mean of population 1 less than the mean of population 2?
\(H_A: \mu_1 <\mu_2\)
Upper one-sided
Question of interest: Is the mean of population 1 greater than the mean of population 2?
\(H_A: \mu_1 >\mu_2\)
Two-sided
Question of interest: Is the mean of population 1 different from the mean of population 2?
\(H_A: \mu_1 \neq\mu_2\)
Motivating Example:
\(H_0:\)\(\mu_A = \mu_B\)
\(H_A:\mu_A \neq \mu_B\)
4. Using the sampled data and the alternative hypothesis, determine what values would be considered “as or more extreme” than the observed sampled statistic.
Lower one-sided: \(H_A: \mu_1<\mu_2\)
Upper one-sided: \(H_A: \mu_1 >\mu _2\)
Two-sided: \(H_A: \mu_1 \neq \mu_2\)



4. Using the sampled data and the alternative hypothesis, determine what values would be considered “as or more extreme” than the observed sampled statistic.
Motivating Example:
\(\overline{x}_A - \overline{x}_B = 4 - 3.5 = 0.5\)
Any difference in sample means greater than 0.5 in magnitude would be considered as or more extreme than the observed sampled statistic.
If the sample sizes are sufficiently large, under the null hypothesis, the distribution of the test statistic used in testing the difference between two population means is a
t distribution with \(\nu\) degrees of freedom.
\(\nu\) represents the Satterthwaite degrees of freedom.
How large do the samples need to be?
If both \(n_1\geq 30\) and \(n_2 \geq 30\), we can move forward.
If either \(n_1 < 30\) or \(n_2 < 30\), we need to look at the sampled distribution(s) of the small sample(s). If there are no clear outliers or strong skewness in the sampled data, we can move forward. If either sampled distribution suggests skewness or outliers, we should not proceed.
Motivating Example:
Are the sample sizes sufficiently large?
Yes, both \(n_1\geq\) and \(n_2\geq30\).
So the null distribution is at distribution with 75.897 degrees of freedom.

When testing the difference in population means, \(\mu_1\) vs. \(\mu_2\), the test statistic is
\[t = \frac{\overline{x}_1-\overline{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}\]
Motivating Example:
\[t = \frac{4-3.5}{\sqrt{\frac{1.5^2}{46} + \frac{1^2}{32}}} = 1.766\]
Lower one-sided:
\(H_A: \mu_1<\mu_2\)
Upper one-sided:
\(H_A: \mu_1>\mu_2\)
Two-sided:
\(H_A: \mu_1\neq \mu_2\)



R code:
pt(t, df)
R code:
1-pt(t, df))
R code:
2*(1-pt(abs(t), df))
where df is the Satterthwaite degrees of freedom, \(\nu\)
Motivating Example:

2*(1-pt(abs(1.766), 75.897)) = 0.0814
Write a 4-part conclusion. The conclusion should be written in the context of the problem and contain the following components:
A statement for the strength of evidence in favor the alternative hypothesis.
Whether to reject or fail to reject the null hypothesis.
The point estimate for the parameter of interest.
A \((1-\alpha)100\%\) confidence interval estimate for the parameter of interest.

Motivating Example: Write a 4-part conclusion with a \(\alpha=0.01\) significance level.
1. There is slightly suggestive evidence that the graduates at University A and University B take a different number of math courses, on average.
2. At the \(\alpha=0.01\) significance level, we fail to reject the null hypothesis.
3. and 4. We are 99% confident that the students at University A take 0.248 fewer to 1.248 more math courses than students at University B on average, with an estimate difference in the average number of math courses of 0.5.
Please locate the Class 15 Activity in Top Hat and answer Questions 1-7.