First Quiz
7/25/2021
1.) 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 \(\bar{x}\) =40.5 hours.
A. Is there evidence to support the claim that battery life exceeds 40 hours? Use α=0.05.
Step 1
Establish Hypothesis
Step 2
Determine Appropriate Statistical Test and Sampling Distribution:
This will be a one-tailed test specifically upper-tailed test. Since the standard deviation of the population is known we will use z-distribution.
\[\LARGE {z_{0}=\frac{\bar{x}- \mu_{0}}{\sigma/\sqrt{n}} }\]
Step 3
State the decision rule
Reject \(H_{0}\) if z>1.645 or the P-value is less than 0.05 .
Computations
\[\LARGE {z_{0}=\frac{40.5-40}{1.25/\sqrt{10}} }\] \[\LARGE {=1.26 }\]Because the test is an upper-tailed test, the P-value is computed as
\[\LARGE {P=1-Φ(z_{0})}\]Locate 1.26 in the z-score to get the \(\phi(z_{0})\) which is 0.
\[\Large {P=1-0.896165}\] \[\Large {=0.103835}\] \[\Large {=0.1038}\]
P-value is greater than 0.05 or z is not greater than 1.645 therefore, we failed to reject the null hypothesis which mean that there is not enough evidence that the battery exceeds 40.5 hours.
B. What is the P-value for the test in part A?
\[\Large {P=1-\phi(z_{0})}=1-\phi(1.26)=1-0.896165=0.103835=0.1038\]or
C. What is the β-error for the text in part B if the true mean life is 42 hours?
Determine \(\bar{x}\) at \(z_{crit}=1.645\)
\[\Large {z=\frac{\bar{x}- \mu_{0}}{\sigma/\sqrt{n}} }\] \[\Large {1.645=\frac{\bar{x}- 40}{1.25/\sqrt{10}} }\] \[\Large {1.645(1.25/\sqrt{10})+40={\bar{x}}}\] \[\Large {\bar{x}}=40.65024334\]
Determine where does \({\bar{x}=40.65024334}\) fall on the distribution if \({\mu_{a}=42}\)
\[\LARGE {z=\frac{40.65024334-42}{1.25/\sqrt{10}} }\] \[\LARGE {=-3.414644256=-3.41} \]
\[\LARGE {\beta}=Φ(-3.41)=0.000325\]
If the true mean is 42 hours, then there is a 0.000325 probability of not rejecting the null hypothesis when it should be rejected.
D. What sample size would be required to ensure that β does not exceed 0.10 if the true mean is 44 hours?
Use the formula in the book
\(\beta\) should not exceed 0.10 therefore \(z_{\beta}=z_{1-0.1}=z_{0.9}=1.282\)
\[\Large {n\simeq\frac{(1.645+1.282)^2(1.25)^2}{(4)^2}}\]
E. Explain how you could answer the question in part A by calculating an appropriate confidence bound on battery life.
Calculating the appropriate confidence bound on battery life leads to getting the interval in which the true value of mean lies. According to the book[1], the test will lead to rejection of null hypothesis if and only if a parameter is not in the interval 100(1-α%)CI[l,u]. So by getting the confidence interval on e battery life, we can know if there is or there is not enough evidence to support the claim that the battery exceeds 40.5 hours. In illustration, the CI for this problem is computed as \((-\infty,\bar{x}+z\frac{\sigma}{\sqrt{n}}]\). Since we are doing an upper-tailed test there is no lower bound. \[\large \bar{x}+z\frac{\sigma}{n}=40.5+1.645\frac{1.25}{\sqrt{10}}=41.15\]
The CI is \(μ ≤ 41.15\). Since \(μ_0 = 40\) is included in this interval, we failed to reject the null hypothesis which is the same result we got in part A.
2. 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).
Follow the steps in the book
1. Parameter of interest: The quantity of interest is the difference in mean mileage, \(μ_1 − μ_2\), and \(Δ_0 = 0\).
2. Null hypothesis: \(H_0: μ_1 − μ_2 = 0,\) or \(H_0: μ_1 = μ_2.\)
3. Alternative hypothesis: \(H_1: μ_1 < μ_2\). We want to reject \(H_0\) if the mileage gasoline B is significantly better than that of A.
4. Test statistic: The test statistic is \[\Large {z=\frac{\bar{x_1}- \bar{x_2}-0}{\sqrt{\frac{\sigma^2_1}{n_1}+\frac{\sigma^2_2}{n_2}} }}\]
5. Reject \(H_0:μ_1 = μ_2\) if: the P-value is less than 0.05.
6. Computations: Because \(x_1\) = 19.6 miles per gallon and \(x_2\) = 20.2 miles per gallon, the test statistic is \[\Large {z=\frac{19.6-20.2}{\sqrt{\frac{0.4^2}{16}+\frac{0.6^2}{16}} }=-3.33}\]
7. Conclusion: Because \(z_0 = -3.33\), the P-value is \(P =Φ(-3.33) = 0.000434\) , \(H_0\) is rejected at the \(α = 0.05\) level.
Practical Interpretation: We conclude that Brand B gasoline is significantly better than A.
References
- D. Montgomery and R. Runger, “Connection between Hypothesis Tests and Confidence Intervals,” in Applied Statistics and Probability for Engineers, 1994, ch.9.1.5, p.206
- B.Foltz, “Statistics 101: Calculating Type II Error, Concept with Example,” Youtube, Mar. 13, 2013. [Video recording]. Available: https://www.youtube.com/watch?v=jxxcwSFq6_Q