Individual Quiz
John Kyle C. Lluz
7/25/2021
Module 3 Quiz
Question 1
The life in hours of a battery is known to be approximately normally distributed with standard deviation of \(\sigma\)= 1.25 hous. A random sample of 10 batteries has an average life of 40.5 hours.
Given:
\[\sigma = 1.25\ Hours \\ n = 10\ batteries \\ \overline{x} = 40.5 \]
a.) Is there evidence to support the claim that the battery life exceeds 40 hours? Use \(\alpha\)=0.05.
1.) Parameter of Interest: The Parameter of interest is \(\mu\), the life hours of the battery
2.) Null Hypothesis: \[H_{o} : \mu = 40 \]
3.) Alternative Hypothesis:
\[H_{\alpha} : \mu > 40 \]
4.) Test Statistic:
\[Z_{o} = \frac{\overline{x}-\mu_{o}}{{\sigma}/{\sqrt{n}}}\]
5.) Reject \(H_{o}\) if:
Reject \(H_{o}\) if \(Z_{o} < Z_{\alpha}\) because there is no significant evidence that the battery life can exceed 40 hours.
6.) Computations:
Finding \(Z_{o}\) :
\[Z_{o} = \frac{\overline{x}-\mu_{o}}{{\sigma}/{\sqrt{n}}}\]
Substitute the Given
\[Z_{o}=\frac{40.5 -40}{{1.25}/{\sqrt{10}}}\]
\[Z_{o}=\frac{0.5}{{1.25}/{3.16227766}}\]
\[Z_{o}=\frac{0.5}{0.3952847075}\]
\[Z_{o}=\frac{0.5}{0.3952847075} = 1.2649 \]
\[Z_{o}= 1.26\]
Finding \(Z_{\alpha}\) :
qnorm(0.05, lower.tail=FALSE)## [1] 1.644854
7.) Conclusions:
\[Z_{o} < Z_{\alpha}\]
\(\therefore\) There is a failure to reject the null hypothesis. There is also not enough evidence to support the claim that the battery life exceeds 40 hours.
b.) What is the P-value in part A?
Formula for P-Value given that it is right-tailed:
\[P(Z \ge z_{o} ) = 1 - \Phi(z_{o})\]
Solve:
\[P(Z \ge z_{o} ) = 1 - \Phi(z_{o})\]
\[P(Z \ge z_{o} ) = 1 - 0.8962\]
\[P(Z \ge z_{o} ) = 0.1038\]
*R has also a function to extract P-values:
1-pnorm(1.26)## [1] 0.1038347
\[Ans: The\ P-value\ in\ Part\ A\ is\ 0.1038\]
c.) What is the \(\beta\)-error for the text in part B if the true mean life is 42 hours?
Formula of \(\beta\)-error
\[\beta = \Phi(Z_{\alpha}-\frac{{\delta}{\sqrt{n}}}{\sigma})\]
\[\beta = \Phi(1.644854-\frac{{2}{\sqrt{10}}}{1.25})\]
\[\beta = \Phi(1.644854-5.059644256)\]
\[\beta = \Phi(-3.41479)\]
d.) What sample size would be required to ensure that \(\beta\) does not exceed 0.10 if the true mean is 44 hours?
Formula to be used:
\[n = \frac{({Z_\alpha}+{Z_\beta})^2 \sigma^2}{\delta^2} \]
\[n = \frac{({1.645}+{1.285})^2 1.25^2}{(44-40)^2} \]
\[n = \frac{({2.93})^2 1.25^2}{(4)^2} \]
\[n = \frac{(8.5849)(1.5625) }{16}\]
\[n = \frac{13.41390625}{16}\]
\[n = 0.8383692406\]
*In reality, you will never have a less than 1 sample size meaning…
\[n = 0.8383692406 \approx 1 \]
\(\therefore\) You will need a sample size of 1 to ensure that the \(\beta\) 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.
Formula of a one-sided interval:
\[ (-\infty \ , \ \overline{x} + z_{a} \frac{\sigma}{\sqrt{n}}) \]
\[ (-\infty \ , \ 40.5 + 1.65 \frac{1.25}{\sqrt{10}}) \]
\[ (-\infty \ , \ 40.5 + (1.65) (0.3952847075)) \]
\[ (-\infty \ , \ 40.5 + 0.6522197674) \]
\[ (-\infty \ , \ 41.15221977) \]
\[ \approx(-\infty \ , \ 41.15) \]
\(\mu = 40\) is included within the confidence interval and that gives the conclusion that there is not enough evidence to reject \(H_o\).
Question 2
Brand A gasoline was used in 16 similar automobiles under identical conditions. The corresponding sample of 16 values (mile per gallon) had a mean of 19.6 and standard deviation of 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 that that of A? Assume normality. Test the hypothesis using both P-values and fixed significance level with \(\alpha=0.05\) approaches (if possible)
Given:
Brand A
\[ \overline{x_1} = 19.6 \]
\[ \delta^2_1 = 0.4 \]
\[ n_1 = 16 \]
Brand B
\[ \overline{x_2} = 20.2 \]
\[ \delta^2_2 = 0.6 \]
\[ n_2 = 16 \]
Significance Level:
\[ \alpha = 0.05 \]
1.) Parameter of Interest
We will look for the difference in the mileage of Gasoline A compared to Gasoline B
2.) Null Hypothesis:
\[H_{o}: \mu_{1} = \mu_{2}\]
3.) Alternative Hypothesis:
\[H_{1}: \mu_{1} < \mu_{2}\]
4.) Test Statistic:
\[T = \frac{{\overline{x_1}}-{\overline{x_2}}-{\Delta_0}}{\sqrt({\frac{\delta^2_1}{n_1}}+{\frac{\delta^2_2}{n_2}})}\]
5.) Reject $ H_{o}$ if:
Reject \(H_o\) if P-value is less than 0.05.
6.) Computations:
\[T = \frac{{\overline{x_1}}-{\overline{x_2}}-{\Delta_0}}{\sqrt({\frac{\delta^2_1}{n_1}}+{\frac{\delta^2_2}{n_2}})}\]
\[T = \frac{19.6-20.2}{\sqrt({\frac{0.4^2}{16}}+{\frac{0.6^2}{16}})}\]
\[T = \frac{-o.6}{\sqrt({\frac{0.16}{16}}+{\frac{0.36}{16}})}\]
\[T = \frac{-o.6}{\sqrt(0.0325)}\]
\[T = \frac{-o.6}{0.1802775638}\]
\[T = \frac{-o.6}{0.1802775638}\]
\[T = -3.3282\]
*using the function in finding the P-Value:
pnorm(-3.3282)## [1] 0.0004370455This then gives us"
\[\approx 0.0004\]
7.) Conclusions:
\(\therefore\) When we compare the P-value to the level of significance, we can say that they are not equal. Because of this, we can reject the null hypothesis and there is enough evidence to support the claim that the mileage of B is significantly better that of A.