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The following table is a cross tabulation of328 people whowatched 3 movies and their age group. There are four groups;under 13, 13-30, 31-60, and 60+. The purpose...

Question

The following table is a cross tabulation of328 people whowatched 3 movies and their age group. There are four groups;under 13, 13-30, 31-60, and 60+. The purpose of this study isto determine if interest in these movies depends on age. AChi-square test of independence is used. What is thechi-square value for this data? Round to 2 decimalsMoviesAge groupTenetGreyhoundWanderGrand Total Under 13122122613-3069138316531-6048105611460+77923Grand Total13632160328

The following table is a cross tabulation of328 people who watched 3 movies and their age group. There are four groups; under 13, 13-30, 31-60, and 60+. The purpose of this study is to determine if interest in these movies depends on age. A Chi-square test of independence is used. What is the chi-square value for this data? Round to 2 decimals Movies Age group Tenet Greyhound Wander Grand Total Under 13 12 2 12 26 13-30 69 13 83 165 31-60 48 10 56 114 60+ 7 7 9 23 Grand Total 136 32 160 328



Answers

Use either the critical-value approach or the P-value approach to perform a chisquare independence test, provided the conditions for using the test are met. Diabetes in Native Americans. Preventable chronic diseases are increasing rapidly in Native American populations. particularly diabetes. F. Gilliland et al. examined the diabetes issue in the paper "Preventative Health Care among Rural American Indians in New Mexico" (Preventative Medicine, Vol. 28. pp. $194-202$ ). Following is a contingency table showing cross classification of educational attainment and diabetic state for a sample of 1273 Native Americans (HS is high school). At the $1 \%$ significance level, do the data provide sufficient evidence to conclude that an association exists between educational level and diabetic state for Native Americans?

All right, so we're gonna be comparing the ratings between Siskel and Ebert. Um No hypothesis is that there is no association between their ratings and alternative hypothesis is that there is an association between their ratings. So I've gone ahead and copied out these values into Excel and here we are plopped them right there, hold up to go fetch it. So here it is. Um what we're looking at here is our originals on the left here, this right here is our original And then this is our expected right here. So we're expecting 17 0.78 So it's going to be about 18, 18 Down Mixed and Up Reviews. So right there and then we take the expected minus D. Observed squared and then divided by the expected to get us these values right here and then we take the sum of all of these values in there to get 146 for our chi square, which results in a p value of basically zero, it's super super small. So we reject the null hypothesis in favor of the alternate

All right. So you're asked about thoughts of suicide and mental health, cheerful topics indeed, At a 5% significance level. See if there is an association. I think we might all know the answer to that right off the bat. But let's prove it statistically. So. I'm just gonna paste out my screenshot from word or exile since that when I used Oh no, There it is. 1374. Here we go. All right. So let's ignore my work from the first part here. That's our calculated values. So response, this is our original table. It's got all our values here. Okay, respondents number, sample size right there. We got four categories in the columns and three in the rows are actually four rows and three columns. So then we can calculate are expected value which is going to be this guy times this guy divided by the sample size and that's gonna get us this value right here and then we proceed in the same way Just making those intersections here and here to get that one this one and this one to get that one And so on. So that's how we get our expected values. We're going to do that for 12 values. And this is why I use excel because I don't want to do this 12 times. Um but then we see that in our expected values um none of them are below five, which is good. So that means we can use archive squared test of independence. Yeah. And now let's remove this scribbles. We find that there are these results here. And then when we add them all up, we get a chi square of 91.25 which is a pretty big chi square statistic. And then for the P value we get a very small number which is basically zero. So at the 5% significance level we reject the null hypothesis that there is no association and accept the alternative hypothesis that there is an association. Mm

Mhm RP have value is going to be given by the number of successes 82 divided by the sample size 328. This is equal to 1/4 or 0.25. RZ critical value for 95% confidence interval. Mhm Is going to be 1.96. Our confidence interval is going to be given by he had plus or minus the Z critical value. Okay. This times the square root of the proportion. Times 1- of proportion. Thank you. Divided by the sample size. Uh huh. Yes, this is approximately equal to 0.21 2.3.

Mhm P hat is going to be given by the sample size. Okay. 328 plus four over the number of successes 82 plus two. This gets us an approximate proportion of 0.253. Mhm At the 595% significance level Or 5% significance level 95% confidence and will is able to be constructed from this C critical value using table too. Yeah. 5% divided by two is 0.025. This is 1.96. To find our 95% confidence interval. We have 253 plus or minus 1.96 times the square roots of the proportion to 5, 3 Times 1 -153 divided by the sample size. This gets us a lower bound of .206, 2.3.


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