The volume of one of the Great Lakes is 3.5 × 10E3 cubic kilometers.
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There are 8 ounces in a cup.
To find how many cups are in a gallon of milk, divide the ounces in a gallon by the ounces in a cup:
A gallon will contain 16 eight-ounce cups.
To find how many cups are in a gallon of milk, divide the ounces in a gallon by the ounces in a cup:
A gallon will contain 16 eight-ounce cups.
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] ± *
Where the margin of error is:
*
=
*
= 4.604*
= 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:
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
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!
