Do seasons influence the type of pet people adopt? a researcher is interested in finding out whether thereis a relationship between the type of pet adopted (dog, cat, bird, etc.) and the season (winter, spring, summer, fall)during which such pet would be adopted. he randomly selects 200 adoption files from the humane society andrecords the season and pet type for each of the 200 adoptions. what kind of test scenario is this? a. simple linear regressionb. one-sample t-test for a population meanc. paired t-test for a population mean differenced. two-sample t-test for the comparison of two population meanse. one-sample z-test for a population proportionf. two-sample z-test for the comparison of two population proportionsg. anova for comparing many population meansh. chi-squared test of goodness of fiti. chi-squared test of independencej. chi-squared test of homogeneityk. one-sample binomial test for a population proportion

Answer = I. Chi-squared test of independence

Step-by-step explanation:

The test of independence uses the contingency table format. The analysis of a two-way contingency table helps to answer the question whether two variables are related or independent of each other. Thus, the chi-square statistic measures how much the observed frequencies differ from the expected frequencies when variables are independent. The first procedure is to state the null and alternative hypothesis  

H0: No relationship exists between the type of pet adopted (dog, cat, bird, etc.) and the season (winter, spring, summer, fall) during which such pet would be adopted  

H1: There is a relationship between the type of pet adopted (dog, cat, bird, etc.) and the season (winter, spring, summer, fall) during which such pet would be adopted  

To be able to decide whether to reject the null hypothesis or not, we need to compare the calculated test statistic with its chi-square table or critical value using a given level of significance and degree of freedom and then decide using the decision rule (Reject H0 if the calculated chi-square statistic is greater than the critical value otherwise, accept H0)

o= 4(i) +3

step-by-step explanation:

o is your output value and i is your input value. so, for example, if we stuck 7 in as our input, we would first multiply by 4 ( pemdas ), which equals 28. then we would add 3 to 28, for a total output of 31. this may not be a graph, but, it is a way to solve your formula..