Understanding Chi-Square Test through Excel Sheet

From the above output, we can see that the p-value is 0.000186, which is smaller than 0.05 (if we select a 0.05 significance level).

This information is true because the data have a p-value of 0.055793. This also allows us to support the null hypothesis, and reject the previously stated hypothesis. Shifting next to hypothesis 2, using the chi squared test to find a p-value of 0.00000000605505 to determine that acidity of a region had a significant relationship between the amount of isopods and the amount of acid or base present. This means that we should reject the null hypothesis, and support our original hypothesis. Now, hypothesis 3 had a p-value of 0.001823 showing us that the relationship between the isopod count and the amount of light present is significant. This information disproves the null hypothesis, as well as the previously stated hypothesis because the isopods were in fact drawn to a darker environment. Lastly, hypothesis 4 gives a p-value of 0.0000000153118 expressing that the relationship between temperature and amount of isopods are significant. This allows us to determine that we should reject the null hypothesis as well as the original hypothesis due to the fact that there was a significant amount of isopods in the cold

17 In regression analysis, the coefficient of determination R2 measures the amount of variation in y

However, we must perform some analysis on this data to confirm that the results for our class really are significantly different from the average. To do this, we performed a chi-square analysis on our data. Our chi-square value is 14.733, and the degrees of freedom are 6. The resulting p-value falls between 0.05 and 0.02. Therefore, we conclude that the data from our class indeed does not fit the average.

(e) Is there a significant relationship between the selling price and the assessed value of the house? Use 5 % level of significance.

Because the p-value of .035 is less than the significance level of .05, I will reject the null hypothesis at 5% level.

1. Are any of the lab values in Table 1 out of normal range? Do you see some that are too high or too

To test the null hypothesis, if the P-Value of the test is less than 0.05 I will reject the null hypothesis.

The next table shows the results of this independent t-test. At the .05 significance level, can we conclude

With a P-value of 0.00, we have a strong level of significance. No additional information is needed to ensure that the data given is accurate.

When thinking about the variables the agency is measuring, a chi-square statistic would help the agency measure the association and agreement for nominal and ordinal data in the intervention, which in this case is employment level and treatment condition. If the treatment condition (intervention) is reliable and effective and shows this was what caused the employment level (outcome) of the participants. Therefore, the agency chose the chi-square statistic to show if there was a relationship between the two variables (independent and dependent). The variables have to be categorical to show by attending the treatment the employment level increased for

This value was calculated to be 0.0166, so the experimental value for π is 3.183 ± 0.0166. Percent error (% error =|(measured-accepted)/accepted|*100%) was then calculated to be 1.4% making the results

The Chi-squared test has certain requirements in terms of the minimum expected population for each category. The data did not meet these criteria, meaning that the chi-squared test in this circumstance would be inappropriate. To avoid such issues, the nonparametric Kolmogorov-Smirnov test was used in preference as it has less rigorous data requirements..

After that, the team had to calculate the test statistic by using the Appendix B worksheet rank sum sample size and rank sum from each group. The following formula gives the results of that worksheet

The instrument used is questionnaire and chi-square is used to test the relationship between the variables, which has proven that there is a