Question

Use a t-distribution to find a confidence interval for the difference in means μd=μ1-μ2 using the relevant sample results from paired data. Assume the results come from random samples from populations that are approximately normally distributed, and that differences are computed using d=x1-x2.

Answer 1

a)

best estimate = Xd[bar]=4.80

margin of error = 8.66

The 99% confidence interval is -3.86 to 13.46

b)

test statistic = -1.86

p-value = 0.0526

Decision: Reject the null hypothesis.

At the 5% significance level, you can conclude that the population mean of the difference between treatment 1 and treatment 2 is less than zero.

Explanation:

Hello!

a) 99% CI

Using d=X₁-X₂ to determine the study variable Xd: the difference between treatment 1 and treatment 2.

Assuming that this variable has an approximately normal distribution: Xd≈N(μd;σ²d)

To calculate the sample mean and standard deviation you have to calculate the difference between the values of both treatments first.

Case 1 ; Case 2 ; Case 3 ; Case 4 ; Case 5

22-18= 4 ; 27-29= -2 ; 32-25= 7; 26-20= 6 ; 29-20= 9

n= 5

Xd[bar]= ∑X/n= 24/5= 4.80

Sd²= 1/(n-1)*[∑X²-(∑X)²/n]= 1/4*[186-(24²)/5]= 17.7

Sd= 4.21

The parameter of interes is the population mean od the difference, μd

The best estimate for this parameter is the sample mean, Xd[bar]=4.80

Using the t-distribution, the formula for the Confidence Interval is

Xd[bar] ±
`t_(n-1;1-\alpha /2)`*
`(Sd)/(√(n) )`

Where the margin of error is:

`t_(n-1;1-\alpha /2)`*
`(Sd)/(√(n) )`=
`t_(4;0.995)`*
`(Sd)/(√(n) )`= 4.604*
`(4.21)/(√(5) )`= 8.66

99% CI [-3.86; 13.46]

b) 5% Hypothesis test

The variable of interest is defined d=X₁-X₂; Xd: the difference between treatment 1 and treatment 2. Xd≈N(μd;σ²d)

The statistic hypotheses are:

H₀: μd = 0

H₁: μd < 0

α: 0.05

The statistic to use for this test is:

`t_(H_0)= (X_d[bar]-Mu_d)/((Sd)/(√(n) ) ) ~~t_(n-1)`

As before you have to calculate the difference between the observation for each case and then the sample mean and standard deviation:

Case 1 ; Case 2 ; Case 3 ; Case 4 ; Case 5 ; Case 6 ; Case 7 ; Case 8

18-18= 0; 12-19= -7; 11-25= -14; 21-21= 0; 15-19= -4; 11-14=-3; 14-15= -1; 22-20= 2

n= 8

Xd[bar]= ∑X/n= -27/8= -3.38

Sd²= 1/(n-1)*[∑X²-(∑X)²/n]= 1/7*[275-(-27²)/8]= 26.27

Sd= 5.13

`t_(H_0)= (-3.38-0)/((5.13)/(√(8) ) )= -1.86`

This test is one-tailed to the left, which means that you will reject the null hypothesis to small values of t, the p-value of the test has the same direction as the rejection region, this means that it is one-tailed to the left and you can calculate it as:

P(≤-1.86)= 0.0526

The decision rule using the p-value is:

If p-value > α, do not reject the null hypothesis.

If p-value ≤ α, reject the null hypothesis.

The p-value is greater than the significance level so the decision is to reject the null hypothesis.

I hope it helps!