Pearson's chi-square test

critical value is a number that separates a region where the null hypothesis will not be rejected from a region where it will be rejected. It is achieved at a particular percentile range and is then determined using a chart, known as the distribution topic.” The equation below states the critical value at 5% significance level: C.95 =21.026 * The .95 is taken from 100%-5%=95% *The 12 is from the degrees of freedom Lastly, I found what conclusion could be drawn regarding the grades achieved on a semester basis and the time spent on a daily basis on social media sites. I used information from stating my null and alternative hypothesis to either accept/reject my null/alternative. I did this through comparing my critical value at 5% and my chi-squared . My outcome is stated below: cal < co (df) = accept the Ho cal > co (df) = reject Ho and accept HI = 54.33 C.95 =21.026…

An excel spreadsheet was provided by the instructor to allow us to fill in information about the assigned dissertations; the following is information required to complete the excel spreadsheet: citation, advisor, advisor’s credentials, purpose, questions, hypothesis, independent and dependent variable(s), sample (n), sampling method, research type, statistical analysis, instrument utilized, reliability and validity, alternate explanation, limitation, future research and finally significance.…

The chi square value is then compared to the p-value, which is a critical value gotten from a chart based on the degree of freedom and percent of error which this experiment used as 0.05. If the chi square value was greater than the critical value, the hypothesis would be rejected, meaning there was too much error and observed results were way off from the expected results. However, if the chi square value was less than or equal to the critical value, then the null hypothesis would not be…

A sample revealed three bags of 56, two bags of 57, one bag of 55, and two bags of 58. To test the hypothesis that the mean candies per bag is 56, how many degrees of freedom are there? * 8 * 9 * 7 * 1 Which of the following is most appropriately displayed with a frequency table? * What percentage of people prefer Hunt's brand ketchup * The home location of the most valuable customers * The relationship between gender and job performance * …

Figure 1. Chi Square Test for Niche lab conducted at Owings Mills North.Behaivors of several species of birds (Tufted titmouse, dark eyed junco, black-capped chickadee, northern cardinal, mourning dove, purple finch, house sparrow, and white breasted nuthatch) were observed at feeding station. The three possible behaviors are feeding at the feeder, feeding on the ground, or feeding in the tree. Assignment questions 1-4 Given the results of the chi square, the birds feeding behavior are…

Chi- Square Goodness of Fit Testing Assuming Equal Expected A handful of students loves numbers and selects a major in Finance, Statistics, or Marketing. While others prefer a major in Leadership or are undecided. Historically, the MBA program at Whatsamatta U has about 40% of their students electing Leadership as their major, 30% selecting Finance as a major, 20% choosing Marketing, and 10% are undecided. Conducting a Chi-Square Goodness of the Fit Test revealed that the class of 200 students…

Results In the lab, 478 trials were conducted using a total of 956 crickets. The 478 pairs resulted in 264 wins for the resident cricket and 214 wins for the intruding crickets. These results produced a chi squared value of 5.230. The p-value calculated from this lab was .022. A chi squared value of .022, less than the benchmark of .05 means the data from the lab was significant and statistically rejects the null hypothesis that there was no correlation between residency and aggression…

White M = 1688.4348, African American M = 1564.6667. The p value (p = .782) is not significant since a significant p value is < .05. Thus there is no significant difference in the mean between white and African American participants accepting the null hypothesis. Analysis of variance (ANOVA) was used to test the hypothesis there is a difference in the mean of IV volume infused by ethnicity. A ratio between groups and within groups reflects an F (5, 93) =.211 which is small suggesting there…

The two variables used for the topic are the level of educational attainment (x) and a health index rating (y). Furthermore, the two simple mathematical processes used for the topic are correlation, and linear regression {possibly}; the advanced mathematical process used for the topic are chi-square test {once learned}. Through the expanding data, will provide a correlation, accomplished through creating a Pearson's product-moment correlation coefficient table, with each summation of seven…

In the Chi-Square test table below, the “Asymp. Sig. (2-tailed)” determines the level of confidence. This particular level of confidence is less than 0.5, which means the difference is statistically significant. The null hypothesis is thus rejected. Chi-Square Tests Value df Asymp. Sig. (2-sided) Pearson Chi-Square 12.747a 6 .047 Likelihood Ratio 13.057 6 .042 Linear-by-Linear Association 5.178 1 .023 N of Valid Cases 742 Correlation and Regression with SPSS Pearson’s r value and…