Hypothesis tests for the normal distribution can be conducted in a very similar way to binomial distribution, except this time we switch our test statistic. These tests are useful as again they help us test claims of normally distributed items.
How do we carry out a hypothesis test for normal distribution?
When we hypothesis test for the mean of a normal distribution we think about looking at the mean of a sample from a population.
So for a random sample of size n of a population, taken from the random variable \(X \sim N(\mu, \sigma^2)\), the sample mean \(\bar{X}\) can be normally distributed by \(\bar{X} \sim N(\mu, \frac{\sigma^2}{n})\).
Let's look at an example.
The weight of crisps is each packet is normally distributed with a standard deviation of 2.5g.
The crisp company claims that the crisp packets have a mean weight of 28g. There were numerous complaints that each crisp packet weighs less than this. Therefore a trading inspector investigated this and found in a sample of 50 crisp packets, the mean weight was 27.2g.
Using a 5% significance level and stating the hypothesis, clearly test whether or not the evidence upholds the complaints.
SOLUTION:
| STEP | Example |
| Step 1: State hypotheses clearly. | \(H_0: \mu = 28 \quad H_1: \mu < 28\) |
| Step 2: Write out the probability distribution assuming H0 is true. | \(X \sim N(28, 2.5^2)\) |
| Step 3: Find the probability distribution of the sample mean. | \(\bar{X} \sim N(28, \frac{2.5^2}{50})\) |
| Step 4: Sketch a normal distribution diagram. |  Sketching normal distribution - StudySmarter Originals
We are going to calculate \(P(\bar{X} \leq 27.1)\). |
| Step 5: Make the calculation \(P(\bar{X} \leq \bar{x})\). \(\bar{x}\) is the mean that was found in the sample. | \(P(\bar{X} \leq 27.1) = 0.00545\) |
| Step 6: Compare with the significance level, and make a conclusion on the hypothesis. | \(0.00545 < 0.05\), and therefore contained within the critical region so we reject our null hypothesis and accept our alternative hypothesis. |
This is an example of a one tailed test. Let's look at an example of a two tailed test.
A machine produces circular discs with a radius R, where R is normally distributed with a mean of 2cm and a standard deviation of 0.3cm.
The machine is serviced and after the service, a random sample of 40 discs is taken to see if the mean has changed from 2cm. The radius is still normally distributed with a standard deviation of 0.3 cm.
The mean is found to be 1.9cm.
Has the mean changed? Test this to a 5% significance level.
| STEP | Example |
| Step 1: State hypotheses clearly. | \(H_0: \mu = 2 \quad H_1: \mu ≠ 2\) |
| Step 2: Write out the probability distribution assuming H0 is true. | \(X \sim N(2, 0.3^2)\) |
| Step 3: Find the probability distribution of the sample mean. | \(\bar{X} \sim N(2, \frac{0.3^2}{40})\) |
| Step 4: Sketch a normal distribution diagram. |  Sketching normal distribution - StudySmarter Originals
|
| Step 5: Make the calculation \(P(\bar{X} \leq \bar{x})\). \(\bar{x}\) is the mean that was found in the sample. | \(P(\bar{X} \leq 1.9) = 0.01751\) |
| Step 6: Compare with the significance level and make a conclusion on the hypothesis. | As this is a two-tailed test we need to divide our significance level by two then compare, so we are comparing with 0.025 not 0.05: \(0.01751 < 0.025\)This is contained within our critical region, therefore we reject our null hypothesis and accept our alternative hypothesis. |
Step 5 may be confusing – do we carry out the calculation with \(P(\bar{X} \leq \bar{x})\) or \(P(\bar{X} \geq \bar{x})\)? As a general rule of thumb if the value is between 0 and the mean, then we use \(P(\bar{X} \leq \bar{x})\). If it is greater than the mean then we use \(P(\bar{X} \geq \bar{x})\).
How about finding critical values and critical regions?
This is the same idea as in binomial distribution. However, in normal distribution, a calculator can make our lives easier.
The distributions menu has an option called inverse normal.
Here, we enter the significance level (Area), the mean (\(\mu\)) and the standard deviation (\(\sigma\)).
The calculator will give us an answer. Let's have a look at an example below.
Wheels are made to measure for a bike. The diameter of the wheel is normally distributed with a mean of 40cm and a standard deviation of 5cm. Some people think that their wheels are too small. Find the critical value of this to a 5% significance level.
SOLUTION
In our calculator, in the inverse normal function, we need to enter:
\(\text{Area} = 0.05 \quad \mu = 40 \quad \sigma = 5\)
The area is 0.05 as the significance level is 5%. This is a one-tailed test – people think that the mean is lower due to the belief that the wheels are too small.If we perform the inverse normal function we get 31.775732.
So that is our critical value and our critical region is \(X \leq 31.775732\).
Let's look at an example with two tails.
The amount of juice in a carton is normally distributed with a mean of 55ml and a standard deviation of 3ml. Customers believe that the mean is actually a different value. SOLUTION This is a two-tailed test so we're going to get two critical values, and a diagram to represent this looks like this:
Two-tailed test - StudySmarter We need to firstly deal with the lower tail. Therefore in our calculator, in the inverse normal function, we need to enter: \(\text{Area} = 0.005 \quad \mu= 55 \quad \sigma = 3\)The area is 0.005 as we need to divide the significance level by two and this is the lower tail (the one 0.005 away from 0). If we perform the inverse normal function we get 47.272512.This is one of our critical values, and our critical region is \(X \leq 47.272512\). Now when we deal with the upper end, we need to change the area in our calculator. This is because we are now 0.005 away from 1 as our area. Therefore:\(\text{Area} = 0.995 \quad \mu = 5 5 \quad \sigma = 3\)
So if we perform our inverse normal function we get 62.727488. This is our other critical value, and our critical region is \(X \geq 62.727488\). This time it is greater than or equal to because it is the upper tail of our two-tailed test. Hypothesis Test for Normal Distribution - Key takeaways
- When we hypothesis test for a normal distribution we are trying to see if the mean is different from the mean stated in the null hypothesis.
- We use the sample mean which is \(\bar{X} \sim N(\mu, \frac{\sigma^2}{n})\).
- In two-tailed tests we divide the significance level by two and test on both tails.
- When finding critical values we use the calculator inverse normal function entering the area as the significance level.
- For two-tailed tests we need to find two critical values on either end of the distribution.
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