The life in hours of a battery is known to be approximately normally distributed with standard deviation σ=1.25 hours. A random sample of 10 batteries has a mean life of x¯=40.5 hours.
A.Is there evidence to support the claim that battery life exceeds 40 hours? Use α=0.05.
B.What is the P-value for the test in part A?
C.What is the β-error for the text in part B if the true mean life is 42 hours?
D.What sample size would be required to ensure that β does not exceed 0.10 if the true mean is 44 hours?
E.Explain how you could answer the question in part A by calculating an appropriate confidence bound on battery life.
Answer:
A. \(z={({40.5}-40})/({{1.25}/{√10})}=1.265\)
We then calculate the p value:
\(Pv=P(Z>1.265)=1-P(Z<1.265)=1-0.897=0.103\)
And since the \(Pv>α\) we have enough evidence to not reject the null hypothesis. Therefore, there is not enough evidence to support the claim that the mean is greater than 40.
B. \(Pv=P(Z>1.265)=1-P(Z<1.265)=1-0.897=0.103\)
C. \(β=P(Z<1.645-(2√10/1.25))=P(Z<-3.409)=0.00033\)
D. In order for the probability of error type II to not exceed 0.1 we can use the following formula:
\(n=(zα+zβ)^2σ^2/(x-μ)^2\)
The true mean is \(μ=44\) and we want to prove that \(β<0.1\) so \(z1-0.1=z0.9=1.29\) represents the value on the normal standard distribution that accumulates 0.1 of the area on the right tail. Then we substitute the values:
\(n={(1.65+1.29)^21.25^2}/(44-40)^2=0.844≈1\)
E. A one sided confidence interval is given by the following solution:
\((-∞,x¯+za(σ/√n))\)
After substituting the values, we get:
\(40.5+1.65(1.25/√10)=41.152\)
So the confidence interval is
\((-∞,41.152)\)
Since 40 is on the confidence interval, we don’t have enough evidence to reject the null hypothesis.
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).
Answer:
Null Hypothesis: μ1=μ2
Alternate Hypothesis: μ2>μ1
\((x¯2-x¯1)/√((σ1^2/n)+(σ2^2/n))\)
\(z=3.33\)
Since \(Pv<0.05\) then we reject the null hypothesis. Since we reject the null hypothesis we can conclude that the mileage of B is significantly better than that of A’s.