The alternative hypothesis: (Ha): μ ≠ 200 Example: Getting a P-Value of 0.000Ī factory claims that they produce tires that each weigh 200 pounds.Īn auditor comes in and tests the null hypothesis that the mean weight of a tire is 200 pounds against the alternative hypothesis that the mean weight of a tire is not 200 pounds, using a 0.05 level of significance. Let’s walk through an example to clear things up. Since 0.000 is lower than all of these significance levels, we would reject the null hypothesis in each case. Common significance levels include 0.1, 0.05, and 0.01. So, when you get a p-value of 0.000, you should compare it to the significance level. If the p-value is less than the significance level, we reject the null hypothesis. That's because, we only have to calculate three of the cells directly.When you run a statistical test, whether it’s a chi-square test, a test for a population mean, a test for a population proportion, a linear regression, or any other test, you’re often interested in the resulting p-value from that test.Ī p-value simply tells you the strength of evidence in support of a null hypothesis. It is in the calculation of the expected values that you can readily see why we have (2−1)(4−1) = 3 degrees of freedom in this case. So, the next thing we need to do is find the expected counts for each cell of the table: Physician The observed data were given to us in the table above. In the context of the two examples above, it quickly becomes apparent that if we wanted to formally test the hypothesis that males and females are distributed equally among the four schools, we'd want to test the hypotheses: Therefore, we can conclude that males and females are not distributed equally among the four schools. and therefore the proportion of students accepted into each school, regardless of gender, is different than the proportion of males and females accepted into each school. If we look at each column in the table, we see that the proportion of males accepted into each school is different than the proportion of females accepted into each school. So, the engineering school also appears to have different rates of acceptance for males and females, 40% compared to 10%.Īgain, there's no need drag this out any further. Of the 800 females who applied to the university, only 80 (or 10%) were accepted into the engineering school. We can see that, of the 1200 males who applied to the university, 480 (or 40%) were accepted into the engineering school. So, the business school appears to have different rates of acceptance for males and females, 20% compared to 30%. And, of the 800 females who applied to the university, 240 (or 30%) were accepted into the business school. In this case, of the 1200 males who applied to the university, 240 (or 20%) were accepted into the business school. Let's again start by focusing on the business school. Therefore, we can conclude that males and females are distributed equally among the four schools. which therefore happens to equal the proportion of students accepted into each school, regardless of gender. If we look at each column in the table, we see that the proportion of males accepted into each school is the same as the proportion of females accepted into each school. We probably don't have to drag this out any further.

So, the engineering school also looks to be in good shape, as an equal percentage of males and females, namely 20%, were accepted into it. Of the 800 females who applied to the university, 160 (or 20%) were accepted into the engineering school. We can see that, of the 1200 males who applied to the university, 240 (or 20%) were accepted into the engineering school. So, the business school looks to be in good shape, as an equal percentage of males and females, namely 25%, were accepted into it. Of the 800 females who applied to the university, 200 (or 25%) were accepted into the business school. We can see that, of the 1200 males who applied to the university, 300 (or 25%) were accepted into the business school.

Let's start by focusing on the business school.

Chi square test minitab