Quiz: Question 2

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Brand A gasoline was used in 16 similar automobiles under identical conditions. The corresponding sample of 16 values (miles per gallon) had mean 19.6 and standard deviation 0.4. Under the same conditions, high-power brand B gasoline gave a sample of 16 values with mean 20.2 and standard deviation 0.6. Is the mileage of B significantly better than that of A? Assume normality, Test the hypothesis using both P-value and fixed significance level with α = 0.05 approaches (if possible).

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Solve this problem using the seven-step procedure.

1. Parameter of Interest: The parameters of interest are the mean mileage for the two brands of gasoline, say μ_A and μ_B, and we are interested in determining whether μ_A - μ_B = 0.

2. Null hypothesis: H₀: μ_A = μ_B

3. Alternative hypothesis: H₁: μ_B > μ_A

4. Test Statistic: The test statistic is

t₀* = ( (x̄₁ - x̄₂ - 0) / √(s₁^2/n₁ + s₂^2/n₂) )

5. Reject H₀ if:

Using P-value:

Reject H₀: μ_A = μ_B if the P-value is less than 0.05.

Using fixed significance level:

Compute the degrees of freedom, v.

#v
mB=20.2; mA=19.6; sB=0.6; sA=0.4; nB=16; nA=16
v<- ( (sB^2/nB + sA^2/nA)^2 )/( (sB^2/nB)^2/(nB-1) + (sA^2/nA)^2/(nA-1) )
v

## [1] 26.13402

Compute the critical value.

#critical value
qt(p=.05, df=v, lower.tail = FALSE)

## [1] 1.705295

Using α = 0.05 and a fixed significance level test, we would reject H₀: μ_A = μ_B if t₀* > 1.705295.

6. Computations:

#t0*
t.test.fromSummaryStats <- function(mu,n,s) 
{
   -diff(mu) / sqrt( sum( s^2/n ) )
}

mu <- c(20.2,19.6)
n <- c(16,16)
s <- c(.6,.4)
t.test.fromSummaryStats(mu,n,s)

## [1] 3.328201

t₀* = 3.328201

#p-value
pt(3.328201, v, lower.tail=FALSE)

## [1] 0.001302689

P-value: P = 0.001302689

7. Conclusion:

Using P-value:

Since the p-value, 0.001302689, is less than 0.05, we reject H₀: μ_A = μ_B at the 0.05 level of significance.

Using fixed significance level:

Since t₀* = 3.328201 > 1.705295, we reject H₀: μ_A = μ_B at the 0.05 level of significance.

Practical Interpretation:

There is strong evidence to conclude that the mileage of B is significantly better than that of A.