Question

For each of the following questions, please identify the appropriate statistical technique to use to answer the scientific question posed. Use the same set of answers for every question. a) 1-sample z test for proportions d) 2-sample z test for proportions g) correlation analysis b) 1-sample t test e) 2-sample t test h) linear regression c) 1-variable chi-square test f) 2-variable chi-square test

Answer 1

19). we use a two sample t test for testing the difference between two independent population mean.if the population means are the same,(μ_{1 -} μ₂₎ will equal zero ,otherwise not zero.where μ₁ is the mean of one population and μ₂ be the mean of another population.

To perform this test, both samples must be normally distributed.to check this i am using minitab for doing a normal probability plot.

steps:-

copy the data in two different columns named 'alcoholic' and 'control' $\rightarrow$ graph $\rightarrow$ probabiility plot $\rightarrow$ multiple $\rightarrow$ in group variables select 'alcoholic' and 'control'$\rightarrow$ok.

Probability Plot of alcoholic, control Normal - 95% CI - Variable alcoholic control E 95+------------ - ---------- --- Mean StDev N ADP 40.13 2.696 8 0.185 0.865 42.13 2.696 8 0.306 0.483 ----- LLLLLLLLL --- --- Percent - . 3035 404550. Data 55

here we see that both samples( alcoholic , control) have a p-value above 0.05 .so, it can be concluded that both samples are normally distributed.hence we can perform a 2 sample t test.

20).

here we will perform 2 sample t test.

because, here the data of January and July are independent on each other.so we can not do paired t test, we have to do 2 sample t test for testing the mean air quality in January and July.

our null hypothesis be:-

Ho: Mjan - Hjuly = 0

H1 : Mjan - Hjuly + 0

If our null hypothesis gets rejected at any specific level of significance then we can say that there is enough evidence to state that air quality varies between winter and summer.

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