Answer:
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.
Step-by-step 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] ± [tex]t_{n-1;1-\alpha /2}[/tex]*[tex]\frac{Sd}{\sqrt{n} }[/tex]
Where the margin of error is:
[tex]t_{n-1;1-\alpha /2}[/tex]*[tex]\frac{Sd}{\sqrt{n} }[/tex]= [tex]t_{4;0.995}[/tex]*[tex]\frac{Sd}{\sqrt{n} }[/tex]= 4.604*[tex]\frac{4.21}{\sqrt{5} }[/tex]= 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:
[tex]t_{H_0}= \frac{X_d[bar]-Mu_d}{\frac{Sd}{\sqrt{n} } } ~~t_{n-1}[/tex]
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
[tex]t_{H_0}= \frac{-3.38-0}{\frac{5.13}{\sqrt{8} } }= -1.86[/tex]
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!
