Professor Grimsby
Statistics teachers are sometimes notorious for using this mathematics on their own students.

He has collected a sample of 33 students which indicate an average of 76.1points.

Test his claim at the 95% confidence level.

(His claim is that µ is significantly > than 74.5)

Null hypothesis = Alternative hypothesis > <

The process of testing these hypotheses is not that straight-forward.
View Hypothesis Testing Flow Chart

We will show by example :

Packages of Hot Chocolate

The packages of hot chocolate that I buy clearly state that they contain 28 grams of product.

In the sampling of 40 packages we find that the mean weight is 27.91 grams and the standard deviation is 0.15 grams. (Is this significantly less, or not? We need hypothesis testing.)

The claim is that there is 28 grams of product as a population mean weight. (The claim is m=28gm.)

We are to test the claim at the 99% confidence level.

The claim is µ =28gm.

Therefore the null hypothesis is also: µ = 28gm.

Next the alternative hypothesis states: µ 28gm. (The 2 statements must be exact logical opposites.)

For the equality(=) hypothesis testing we apply our testing on the Left Tail and the Right Tail of the normal distribution. This is always a 2-sided test.

The 2 critical z-values are found in standard normal tables as z = -2.575 and z = + 2.575.

(The table is used to search value closest to .4950 so that 0.5% of the area is in each tail.)

The 2 critical z-values may or may not contain our test statistic.

Out test statistic is calculated by formula:





z = (27.91 - 28.00 ) / 0.15/40 = -3.79

Is the test statistic in one of the critical regions beyond the values of z= -2.575 or z = +2.575 ?

Yes, z= -3.79 is to the left of critical value z = -2.575 and is in the critical region.

In the critical region means we reject the null hypothesis that: µ= 28grams.

Therefore we reject claim that there is a population mean weight of 28grams per package.

Solution:

The claim is clearly that : µ > 74.5 ( No equality so it can't be a null hypothesis.)

Null and alternative hypotheses must be exact logical opposites:

Null hypothesis is : µ 74.5
Alternative hypothesis: µ > 74.5 (This is the claim.)

One-side test or 2-sided test? It is a one-sided test on the Right-tail side.

We need one critical z-value. Since we need 95% confidence we look for 0.4500 in the standard Normal Table and we find the critical z = +1.645 (right side)

Our test statistic is the z calculation where

Test statistic z = (76.1 - 74.5) / (12.8/331/2) = 0.7181 ( a relatively small z-value.)

Is it in the critical region? The critical region starts to the right of z = 1.645.

Our test statistic is not to the right of the critical value. It is not in the critical region.

The null hypothesis is not rejected. Therefore we fail to reject the null hypothesis.

The claim in this question is the same as the alternative hypothesis.
The claim of population mean being greater than 74.5 is not strongly accepted.

Sorry, Professor Grimsby we do not agree with your claim.

Last note:

For a statistician this is the worst scenario. There is no strong conclusion.
If the null hypothesis is rejected than the alternative hypothesis is strongly accepted.

But, failing to reject the null hypothesis seems very much a failure--back to another hypothesis?

Testing a claim about a mean when a small sample is available.

In the last two examples we used the standard normal table as well as z-values for critical values.

For a small sample you may remember that Gosset invented another distribution called the t-distribution. ( it uses t-values and uses test statistic )

Canadian dimes

The easiest explanation of the weight of a Canadian dime is that it weighs 2 grams.
But is this actually a reality or are there many variations?

The first 10 dimes that I weighted in the Science lab came up as follows:
(scale accurate to 0.0001 gram.)

We will claim that the population mean weight for a Canadian dime is greater than 2.0000 grams

Test at 95% confidence level.

Solution:

The 10 weights of the sample provide a sample mean of 1.98356gm and a sample standard deviation of 0.1560gm.

Claim is : µ > 2.000 ( Canadian dimes should be greater than 2.000gm ? )

Null hypothesis is: µ 2.000
Alternative hypothesis is: µ > 2.000

One-sided test or 2-sided test? It is a one-sided Right-Tail test.

Critical t-value is found in t-distribution Table so that t = +1.833
(Using (10-1)=9 degrees of freedom.)

Our test statistic is calculated

t = (1.98356 - 2.0000) / (0.1560/ 101/2 ) = -0.33 (small t-value and on Left side.)

This test statistic is not in the critical region. We fail to reject the null hypothesis.
We fail to accept the alternative hypothesis.

Our sample (n=10) is a very small one but it does not indicate that the mean of all Canadian dimes should weigh more than 2.000 grams.

We say that we are 95% confident ( - E) < m < (+ E ).

Now we investigate statements or claims about the unknown population mean and draw certain conclusions.

This process will be called hypothesis testing.

Definitions and Fundamentals

A hypothesis is a claim or a statement about a parameter of the population. (In this chapter we focus on the population mean parameter .)

The null hypothesis is a statement about the value of a population parameter such that it must contain the condition of equality. ( We use = or or for null the hypothesis.)

The alternative hypothesis is the statement that must be true if the null hypothesis is false.
(We use or > or < for the alternative hypothesis.)