It’s A Math, Math World (Hypothesis Tests))

Today’s blog post is really targeted as a “Statistics 101” for the early statistics student or non-statistician. Last week, we summarized the Central Limit Theorem and Confidence Intervals. This week we look at more inferential statistics, particularly hypothesis tests. Let me know your opinions!

Another way to see if 2 population means differ is to use a hypothesis test. To begin, we formulate a statement, or hypothesis, that both means do not differ and use the sample data to determine whether the collected data is consistent or inconsistent with this hypothesis.

We define the Null hypothesis as H0: u1 = u2 (no difference)

And the Alternative hypothesis as H1: u1 ≠ u2 (two-tailed test)

The corresponding one-tailed tests would yield H1: u1 < u2   OR

                                                                                                      u1 > u2          

We will consider the two tailed test in this case.

We have to determine the test statistic which is the calculated value that we will use to test the null hypothesis.  In this case, we will assume the H0 is true:

t-stat = [(ybar1 – ybar2) – (u1-u2)] /SE (ybar1-ybar2)

We also have the significance level of the test that determines the degree of uncertainty or error in our process.   This significance level, in our case it is equal to 0.05, determines, through a probability table, a certain critical value that we compare the t-stat to in order to determine whether to accept or reject the H0.

For example if |t-stat| > critical value, we say that the t-stat falls in the critical or rejection region and we reject the null hypothesis and conclude that the means are not equal.

However, if |t-stat| < critical value, then the t-stat is not in the critical region and we fail to reject the H0 and we conclude that there is not enough evidence to reject the H0.

Example: 

In this case, we are considering two independent samples of size ≥ 30 so that we may use the standard normal distribution tables (smaller sample sizes would use the students-t distribution)

α = level of significance = 0.05

A study was conducted on two engines and it was hoped that the new engine produced less noise than their old engine. We want to test whether there is any difference between the noise levels produced by the engines.

SAMPLE SAMPLE SIZE SAMPLE MEAN (DECIBELS) SAMPLE STANDARD DEVIATION (DECIBELS)
STANDARD MODEL 65 84 11.6
NEW MODEL 41 72 9.2

We define the Null hypothesis as H0: u1 = u2 (no difference)

And the Alternative hypothesis as H1: u1 ≠ u2 (two-tailed test)

Let:

xbar1 = 84

xbar2 = 72

u1 – u2 = 0 (we assume H0 is true)

s1 = 11.6

s2 = 9.2

n1 = 65

n2 = 41

SE = sqrt((s1)2/n1 + (s2)2/n2) = sqrt ((11.6)2/65 + (9.2)2/41) = sqrt (2.0702 + 2.0644) = sqrt(4.1646)

                = 2.03

T-stat = [(xbar1 – xbar2) – (u1-u2)]/SE

                = (84-72 -0)/2.03

                = 5.91

The Standard normal critical value is z0.025 in each tail of the distribution. This value, read from the Standard Normal Table is 1.96

Since 5.91 > 1.96, the t-statistic falls in the critical (rejection) region for the H0.

Conclusion: We reject the null hypothesis and conclude that there is a difference in noise output between the 2 engines.

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  1. Hey! Is it OK if I go a bit off topic? I am trying to view your domain on my Macbook but it doesn’t display properly, any suggestions? Thanks in advance! Annita

  2. Problem is easy. i have a doubt in the table value of z. Because since it’s one tailed test and level of significance is 0.05 table value of z is 1.645. pls check.any way its not going to affect the final answer. but still

  3. Nicely done. I’ve enjoyed your posts in this series and look forward to more. Your clear simple style helps refresh what I’ve forgotten over the last 40 years.

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