Question

Just for part b. Please do the part b thank u!

1. In a study of egg cell maturation, the eggs from each of four female frogs were divided into two batches and one batch was exposed to progesterone. After two minutes, the cAMP content was measured and the data is tabulated below. It is believed that cAMP is a substance that can mediate cellular response to hormones. Frogotrol progesterone diff -1 3 (a) Use a Randomization Test to test whether there is a mean difference in cAMP between the control group and progesterone group. (Hint: It is a paired data. Under the nul hypothesis, within each pair, control and progesterone are exchangeable.) (b) Perform the usual parametric procedure for the same hypotheses as part a and compare the P values and conclusions from these two approaches. Also list the assumptions you had to make for this parametric approach

Answer 1

Part a)

let xi denotes the cAMP values under control and yi denotes the cAMP values under progesterone.

let us define di=xi-yi for all i=1to n here n=4

We want to test

H0: µd=0 against H1:µ$\neq$0

The differences are 2,-1,3,2 and the sample size is 4

There can be 2⁴ =16 possible differences for each difference signs being minus or plus due to random assignments

The 16 possible differences will be

(2,-1,3,2),(-2,-1,3,2),(2,1,3,2),(-2,1,3,2),(2,1,-3,2),(-2,1,-3,2),(2,-1,-3,2)(-2,-1,-3,2),(2,1,3,-2),(-2,1,-3,-2),(2,-1,3,-2)(-2,-1,3,-2),(2,-1,-3,-2),(-2,-1,-3,-2),(2,1,-3,-2),(-2,1,-3,-2)

here we consider the test statistic $\overline{d}$ for randomisation test

For the original dataset $\overline{d}$=1.5

The no. of cases where calculated $\left | \bar{d} \right |$ from the 16 possible differences> observed $\overline{d}$ is 4

P( no. of $\left | \bar{d} \right |$ > observed $\overline{d}$ )=4/16=1/4=.25

This is the p-value of the test which is .25>.05

so we accept the null hypothesis at .05 level of significance.

Part b)

Parametric procedure:

Assumptions: cAMP content is a variable which is continuous.

We assume that the variable is normally distributed.

let xi denotes the cAMP values under control and yi denotes the cAMP values under progesterone.

let us define zi=xi-yi for all i=1to n here n=4

We want to test

H0: µz=0 against H1:µz$\neq$0

We define the test statistic

t0=$\bar{z}$/(s.d(z)/$\sqrt{n}$)

so here t0=2*1.5/1.732051=2*.866025=1.732051

we reject the null hypothesis at $\alpha$= .05 if

to
> t $\alpha /2$,n-1=t.025,3=3.182

so we accept the null hypothesis

the p-value of the test is 2*p(T>t0)=2*p(T>1.732051)>.05 where T~t with n-1 degrees of freedom

so we accept the null hypothesis. so whether it is randomisation test or parametric test we accept the null hypothesis at .05 level of significance.

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