how to find null hypothesis in minitab

8.1.2.2 - Minitab: Hypothesis Tests for One Proportion

A hypothesis test for one proportion can be conducted in Minitab. This can be done using raw data or summarized data.

• If you have a data file with every individual's observation, then you have  raw data .
• If you do not have each individual observation, but rather have the sample size and number of successes in the sample, then you have summarized data.

The next two pages will show you how to use Minitab to conduct this analysis using either raw data or summarized data .

Note that the default method for constructing the sampling distribution in Minitab is to use the exact method.  If \(np_0 \geq 10\) and \(n(1-p_0) \geq 10\) then you will need to change this to the normal approximation method.  This must be done manually.  Minitab will use the method that you select, it will not check assumptions for you!

8.1.2.2.1 - Minitab: 1 Proportion z Test, Raw Data

If you have data in a Minitab worksheet, then you have what we call "raw data."  This is in contrast to "summarized data" which you'll see on the next page.

In order to use the normal approximation method both \(np_0 \geq 10\) and \(n(1-p_0) \geq 10\). Before we can conduct our hypothesis test we must check this assumption to determine if the normal approximation method or exact method should be used. This must be checked manually.  Minitab will not check assumptions for you.

In the example below, we want to know if there is evidence that the proportion of students who are male is different from 0.50.

\(n=226\) and \(p_0=0.50\)

\(np_0 = 226(0.50)=113\) and \(n(1-p_0) = 226(1-0.50)=113\)

Both \(np_0 \geq 10\) and \(n(1-p_0) \geq 10\) so we can use the normal approximation method. 

Minitab ®  – Conducting a One Sample Proportion z Test: Raw Data

Research question:  Is the proportion of students who are male different from 0.50?

• class_survey.mpx
• In Minitab, select Stat > Basic Statistics > 1 Proportion
• Select One or more samples, each in a column from the dropdown
• Double-click the variable  Biological Sex  to insert it into the box
• Check the box next to  Perform hypothesis test and enter  0.50  in the  Hypothesized proportion  box
• Select Options
• Use the default  Alternative hypothesis  setting of  Proportion ≠ hypothesized proportion value 
• Use the default  Confidence level  of 95
• Select  Normal approximation method
• Click OK and OK

The result should be the following output:

Event: Biological Sex = Male p: proportion where Biological Sex = Male Normal approximation is used for this analysis.

Summary of Results

We could summarize these results using the five-step hypothesis testing procedure:

\(np_0 = 226(0.50)=113\) and \(n(1-p_0) = 226(1-0.50)=113\) therefore the normal approximation method will be used.

 \(H_0\colon p = 0.50\)

 \(H_a\colon p \ne 0.50\)

From the Minitab output, \(z\) = -1.86

From the Minitab output, \(p\) = 0.0625

\(p > \alpha\), fail to reject the null hypothesis

There is NOT enough evidence that the proportion of all students in the population who are male is different from 0.50.

8.1.2.2.2 - Minitab: 1 Sample Proportion z test, Summary Data

Example: overweight.

The following example uses a scenario in which we want to know if the proportion of college women who think they are overweight is less than 40%. We collect data from a random sample of 129 college women and 37 said that they think they are overweight.

First, we should check assumptions to determine if the normal approximation method or exact method should be used:

\(np_0=129(0.40)=51.6\) and \(n(1-p_0)=129(1-0.40)=77.4\) both values are at least 10 so we can use the normal approximation method.

Minitab ®  – Performing a One Proportion z Test with Summarized Data

To perform a one sample proportion  z  test with summarized data in Minitab:

• Select Summarized data from the dropdown
• For number of events, add 37 and for number of trials add 129.
• Check the box next to  Perform hypothesis test and enter  0.40  in the  Hypothesized proportion  box
• Use the default  Alternative hypothesis  setting of  Proportion < hypothesized proportion value 

Event: Event proportion Normal approximation is used for this analysis.

\(H_0\colon p = 0.40\)

\(H_a\colon p < 0.40\)

From output, \(z\) = -2.62

From output, \(p\) = 0.004

\(p \leq \alpha\), reject the null hypothesis

There is evidence that the proportion of women in the population who think they are overweight is less than 40%.

8.1.2.2.2.1 - Minitab Example: Normal Approx. Method

Example: gym membership.

Research question:  Are less than 50% of all individuals with a membership at one gym female?

A simple random sample of 60 individuals with a membership at one gym was collected. Each individual's biological sex was recorded. There were 24 females. 

First we have to check the assumptions:

  np = 60 (0.50) = 30

  n(1-p) = 60(1-0.50) = 30

The assumptions are met to use the normal approximation method.

• For number of events, add 24 and for number of trials add 60.

\(np_0=60(0.50)=30\) and \(n(1-p_0)=60(1-0.50)=30\) both values are at least 10 so we can use the normal approximation method.

\(H_0\colon p = 0.50\)

\(H_a\colon p < 0.50\)

From output, \(z\) = -1.55

From output, \(p\) = 0.061

\(p \geq \alpha\), fail to reject the null hypothesis

There is not enough evidence to support the alternative that the proportion of women memberships at this gym is less than 50%.

Statistics Made Easy

Understanding the Null Hypothesis for ANOVA Models

A one-way ANOVA is used to determine if there is a statistically significant difference between the mean of three or more independent groups.

A one-way ANOVA uses the following null and alternative hypotheses:

• H 0 :  μ 1  = μ 2  = μ 3  = … = μ k  (all of the group means are equal)
• H A : At least one group mean is different   from the rest

To decide if we should reject or fail to reject the null hypothesis, we must refer to the p-value in the output of the ANOVA table.

If the p-value is less than some significance level (e.g. 0.05) then we can reject the null hypothesis and conclude that not all group means are equal.

A two-way ANOVA is used to determine whether or not there is a statistically significant difference between the means of three or more independent groups that have been split on two variables (sometimes called “factors”).

A two-way ANOVA tests three null hypotheses at the same time:

• All group means are equal at each level of the first variable
• All group means are equal at each level of the second variable
• There is no interaction effect between the two variables

To decide if we should reject or fail to reject each null hypothesis, we must refer to the p-values in the output of the two-way ANOVA table.

The following examples show how to decide to reject or fail to reject the null hypothesis in both a one-way ANOVA and two-way ANOVA.

Example 1: One-Way ANOVA

Suppose we want to know whether or not three different exam prep programs lead to different mean scores on a certain exam. To test this, we recruit 30 students to participate in a study and split them into three groups.

The students in each group are randomly assigned to use one of the three exam prep programs for the next three weeks to prepare for an exam. At the end of the three weeks, all of the students take the same exam. 

The exam scores for each group are shown below:

When we enter these values into the One-Way ANOVA Calculator , we receive the following ANOVA table as the output:

Notice that the p-value is 0.11385 .

For this particular example, we would use the following null and alternative hypotheses:

• H 0 :  μ 1  = μ 2  = μ 3 (the mean exam score for each group is equal)

Since the p-value from the ANOVA table is not less than 0.05, we fail to reject the null hypothesis.

This means we don’t have sufficient evidence to say that there is a statistically significant difference between the mean exam scores of the three groups.

Example 2: Two-Way ANOVA

Suppose a botanist wants to know whether or not plant growth is influenced by sunlight exposure and watering frequency.

She plants 40 seeds and lets them grow for two months under different conditions for sunlight exposure and watering frequency. After two months, she records the height of each plant. The results are shown below:

In the table above, we see that there were five plants grown under each combination of conditions.

For example, there were five plants grown with daily watering and no sunlight and their heights after two months were 4.8 inches, 4.4 inches, 3.2 inches, 3.9 inches, and 4.4 inches:

She performs a two-way ANOVA in Excel and ends up with the following output:

We can see the following p-values in the output of the two-way ANOVA table:

• The p-value for watering frequency is 0.975975 . This is not statistically significant at a significance level of 0.05.
• The p-value for sunlight exposure is 3.9E-8 (0.000000039) . This is statistically significant at a significance level of 0.05.
• The p-value for the interaction between watering  frequency and sunlight exposure is 0.310898 . This is not statistically significant at a significance level of 0.05.

These results indicate that sunlight exposure is the only factor that has a statistically significant effect on plant height.

And because there is no interaction effect, the effect of sunlight exposure is consistent across each level of watering frequency.

That is, whether a plant is watered daily or weekly has no impact on how sunlight exposure affects a plant.

Additional Resources

The following tutorials provide additional information about ANOVA models:

How to Interpret the F-Value and P-Value in ANOVA How to Calculate Sum of Squares in ANOVA What Does a High F Value Mean in ANOVA?

Featured Posts

Hey there. My name is Zach Bobbitt. I have a Masters of Science degree in Applied Statistics and I’ve worked on machine learning algorithms for professional businesses in both healthcare and retail. I’m passionate about statistics, machine learning, and data visualization and I created Statology to be a resource for both students and teachers alike.  My goal with this site is to help you learn statistics through using simple terms, plenty of real-world examples, and helpful illustrations.

2 Replies to “Understanding the Null Hypothesis for ANOVA Models”

Hi, I’m a student at Stellenbosch University majoring in Conservation Ecology and Entomology and we are currently busy doing stats. I am still at a very entry level of stats understanding, so pages like these are of huge help. I wanted to ask, why is the sum of squares (treatment) for the one way ANOVA so high? I calculated it by hand and got a much lower number, could you please help point out if and where I went wrong?

As I understand it, SSB (treatment) is calculated by finding the mean of each group and the grand mean, and then calculating the sum of squares like this: GM = 85.5 x1 = 83.4 x2 = 89.3 x3 = 84.7

SSB = (85.5 – 83.4)^2 + (85.5 – 89.3)^2 + (85.5 – 84.7)^2 = 18.65 DF = 2

I would appreciate any help, thank you so much!

Hi Theo…Certainly! Here are the equations rewritten as they would be typed in Python:

### Sum of Squares Between Groups (SSB)

In a one-way ANOVA, the sum of squares between groups (SSB) measures the variation due to the interaction between the groups. It is calculated as follows:

1. **Calculate the group means**: “`python mean_group1 = 83.4 mean_group2 = 89.3 mean_group3 = 84.7 “`

2. **Calculate the grand mean**: “`python grand_mean = 85.5 “`

3. **Calculate the sum of squares between groups (SSB)**: Assuming each group has `n` observations: “`python n = 10 # Number of observations in each group

ssb = n * ((mean_group1 – grand_mean)**2 + (mean_group2 – grand_mean)**2 + (mean_group3 – grand_mean)**2) “`

### Example Calculation

For simplicity, let’s assume each group has 10 observations: “`python n = 10

ssb = n * ((83.4 – 85.5)**2 + (89.3 – 85.5)**2 + (84.7 – 85.5)**2) “`

Now calculate each term: “`python term1 = (83.4 – 85.5)**2 # term1 = (-2.1)**2 = 4.41 term2 = (89.3 – 85.5)**2 # term2 = (3.8)**2 = 14.44 term3 = (84.7 – 85.5)**2 # term3 = (-0.8)**2 = 0.64 “`

Sum these squared differences: “`python sum_of_squared_diffs = term1 + term2 + term3 # sum_of_squared_diffs = 4.41 + 14.44 + 0.64 = 19.49 ssb = n * sum_of_squared_diffs # ssb = 10 * 19.49 = 194.9 “`

So, the sum of squares between groups (SSB) is 194.9, assuming each group has 10 observations.

### Degrees of Freedom (DF)

The degrees of freedom for SSB is calculated as: “`python df_between = k – 1 “` where `k` is the number of groups.

For three groups: “`python k = 3 df_between = k – 1 # df_between = 3 – 1 = 2 “`

### Summary

– **SSB** should consider the number of observations in each group. – **DF** is the number of groups minus one.

By ensuring you include the number of observations per group in your SSB calculation, you can get the correct SSB value.

Leave a Reply Cancel reply

Your email address will not be published. Required fields are marked *

Join the Statology Community

Sign up to receive Statology's exclusive study resource: 100 practice problems with step-by-step solutions. Plus, get our latest insights, tutorials, and data analysis tips straight to your inbox!

By subscribing you accept Statology's Privacy Policy.

Using the t-value to determine whether to reject the null hypothesis

To determine whether to reject the null hypothesis using the t-value, compare the t-value to the critical value. The critical value is t α/2, n–p-1 , where α is the significance level, n is the number of observations in your sample, and p is the number of predictors.

If the absolute value of the t-value is greater than the critical value, you reject the null hypothesis. If the absolute value of the t-value is less than the critical value, you fail to reject the null hypothesis. You can calculate the critical value in Minitab or find the critical value from a t-distribution table in most statistics books. For more information calculating the critical value in Minitab, go to Using the inverse cumulative distribution function (ICDF) and click Use the ICDF to calculate critical values .

• Minitab.com
• License Portal
• Cookie Settings

You are now leaving support.minitab.com.

Click Continue to proceed to:

• Quality Improvement
• Talk To Minitab

One-Sample t-test: Calculating the t-statistic is not really a bear

Topics: Hypothesis Testing , Data Analysis , Statistics

While some posts in our Minitab blog focus on understanding t-tests and t-distributions this post will focus more simply on how to hand-calculate the t-value for a one-sample t-test (and how to replicate the p-value that Minitab gives us). 

The formulas used in this post are available within Minitab Statistical Software by choosing the following menu path: Help > Methods and Formulas > Basic Statistics > 1-sample t .

The null and three alternative hypotheses for a one-sample t-test are shown below:

The default alternative hypothesis is the last one listed: The true population mean is not equal to the mean of the sample, and this is the option used in this example.

For this example, we will use column C2, titled Age, in the Bears.MTW data set, and we will test the hypothesis that the average age of bears is 40. First, we’ll use Stat > Basic Statistics > 1-sample t to test the hypothesis:

After clicking OK above we see the following results in the session window:

With a high p-value of 0.361, we don’t have enough evidence to conclude that the average age of bears is significantly different from 40. 

Now we’ll see how to calculate the T value above by hand.

The formula for the T value (0.92) shown above is calculated using the following formula in Minitab:

The output from the 1-sample t test above gives us all the information we need to plug the values into our formula:

Sample mean: 43.43

Sample standard deviation: 34.02

Sample size: 83

We also know that our target or hypothesized value for the mean is 40.

Using the numbers above to calculate the t-statistic we see:

t = (43.43-40)/34.02/√83) = 0.918542 (which rounds to 0.92, as shown in Minitab’s 1-sample t-test output)

Now, we could dust off a statistics textbook and use it to compare our calculated t of 0.918542 to the corresponding critical value in a t-table, but that seems like a pretty big bear to wrestle when we can easily get the p-value from Minitab instead.  To do that, I’ve used Graph > Probability Distribution Plot > View Probability :

In the dialog above, we’re using the t distribution with 82 degrees of freedom (we had an N = 83, so the degrees of freedom for a 1-sample t-test is N-1).  Next, I’ve selected the Shaded Area tab:

In the dialog box above, we’re defining the shaded area by the X value (the calculated t-statistic), and I’ve typed in the t-value we calculated in the X value field. This was a 2-tailed test, so I’ve selected Both Tails in the dialog above.

After clicking OK in the window above, we see:

We add together the probabilities from both tails, 0.1805 + 0.1805 and that equals 0.361 – the same p-value that Minitab gave us for the 1-sample t test. 

That wasn’t so bad—not a difficult bear to wrestle at all!

You Might Also Like

• Trust Center

© 2023 Minitab, LLC. All Rights Reserved.

• Terms of Use
• Privacy Policy
• Cookies Settings