A scenario where you could use the Chi square goodness of fit test OR the Chi square test of independence Set up the test in SPSS and find out at what sample size the differences between groups become significant or at what counts per cell in the Chi square test of independence the results become significant The point here is to get practice running one of the Chi square methods and to see what relationship there is between sample size and the proportions you are comparing For example, if we go into a typical online classroom of 20 people and ask people whether they prefer one of three pizza toppings (cheese only, vegetables only, meat only), we might propose to expect an even distribution of scores, so 33 per category At 20 people, this is just over 6 people per group, so we might expect frequencies of Cheese (6 people), Vegetable (6 people), and Meat (8 people) We would then compare our actual distribution of people's responses to these expectations using SPSS or an online Chi square goodness of fit calculator to see if the counts per cell (frequencies or proportions) do or don't match the expected distribution of responses If our observed data is Cheese(n 3), Vegetable (n 3), and Meat (n 14), the Chi square value is The Chi 2 value is 7 5 The p value is 02352 The result is significant at p 05 Try out a few of your own calculations to see what size difference between the observed and expected frequencies you need before the results are statistically significant Use samples of 20 30 people and list your expected frequencies, observed frequencies and Chi square values List one example of a non significant result and one example of a significant result Please attach the SPSS data and with response Discuss the observations or question how could you use this, what was clear, what was unclear You may also want to try out these calculators in addition to SPSS https www socscistatistics com tests goodnessoffit default2 aspx https www socscistatistics com tests chisquare2 default2 aspx

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A scenario where you could use the Chi-square goodness of fit test OR the Chi-square test of independence.
Set up the test in SPSS and find out at what sample size the differences between groups become significant or at what counts per cell in the Chi-square test of independence the results become significant.

The point here is to get practice running one of the Chi-square methods and to see what relationship there is between sample size and the proportions you are comparing. For example, if we go into a typical online classroom of 20 people and ask people whether they prefer one of three pizza toppings (cheese only, vegetables only, meat only), we might propose to expect an even distribution of scores, so 33% per category.
At 20 people, this is just over 6 people per group, so we might expect frequencies of Cheese (6 people), Vegetable (6 people), and Meat (8 people). We would then compare our actual distribution of people's responses to these expectations using SPSS or an online Chi-square goodness of fit calculator to see if the counts per cell (frequencies or proportions) do or don't match the expected distribution of responses. If our observed data is Cheese(n=3), Vegetable (n=3), and Meat (n=14), the Chi-square value is: The Chi^2 value is 7.5. Thep-value is .02352. The result is significant atp< .05.

Try out a few of your own calculations to see what size difference between the observed and expected frequencies you need before the results are statistically significant.Use samples of 20-30 people and list your expected frequencies, observed frequencies and Chi-square values. List one example of a non-significant result and one example of a significant result.
Please attach the SPSS data and with response
Discuss the observations or question - how could you use this, what was clear, what was unclear.

You may also want to try out these calculators in addition to SPSS

https://www.socscistatistics.com/tests/goodnessoffit/default2.aspx

https://www.socscistatistics.com/tests/chisquare2/default2.aspx