Individual Quiz
Gian Villamayor
7/22/2021
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.
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A. Is there evidence to support the claim that battery life exceeds 40 hours? Use α = 0.05.
PARAMETER OF INTEREST:
Mean life of battery (\(\mu\))
NULL HYPOTHESIS:
\(H_0:\:\mu =40\:hours\)
ALTERNATIVE HYPOTHESIS:
\(H_1:\:\mu >40\:hours\)
TEST STATISTIC:
\(z_0=\frac{\bar{x}-\mu _0}{\frac{\sigma }{\sqrt{n}}}\)
REJECT \(H_0\) IF:
Reject the null hypothesis if the P-value is less than α (0.05).
COMPUTATION:
Given:
| Statistical Figures | Value |
|---|---|
| \(\bar{x}\) | 40.5 hours |
| σ | 1.25 hours |
| \(\mu _0\) | 40 hours |
| n | 10 |
\(z_0=\frac{40.5\:-\:40}{\frac{1.25}{\sqrt{10}}}\)
\(z_0=1.265\)
CONCLUSION:
We can calculate the p-value using the upper-tailed test since \(H_0:\:\mu =\mu _0\) and \(H_1:\:\mu >\mu _0\)
P = [1 - \(\phi\)(1.265)]
P = (1 - 0.8971)
P = 0.1029
We fail to reject the null hypothesis since the P-value is greater than 0.05. Therefore, there is no evidence to support the claim that battery life exceeds 40 hours.
B. What is the P-value for the test in part A?
Since we are testing \(H_0:\:\mu =40\:hours\) and \(H_1:\:\mu >40\:hours\), we use the upper-tailed test. Therefore, the P-value is computed as:
P = [1 - \(\phi\)(1.265)]
P = (1 - 0.8971)
P = 0.1029
C. What is the β-error for the text in part B if the true mean life is 42 hours?
In order to find the β-error, we will use the formula:
\(\beta =\phi \left(z_{\alpha }-\frac{\delta \sqrt{n}}{\sigma }\right)\)
The critical value for the upper-tailed test with α = 0.05 is 1.645. Therefore we have:
| Statistical Figures | Value |
|---|---|
| \(z_{\alpha }\) | 1.645 |
| \(\delta\) | 2 |
| \(\sigma\) | 1.25 |
| n | 10 |
\(\beta =\phi \left(1.645-\frac{2\sqrt{10}}{1.25}\right)\)
\(\beta =\phi \left(-3.4146\right)\)
β = 0.0.0003
D. What sample size would be required to ensure that β does not exceed 0.10 if the true mean is 44 hours?
In order to get the sample size that will ensure that β does not exceed 0.10, we will use the formula:
\(n=\frac{\left(z_{\alpha }+z_{\beta }\right)^2\sigma ^2}{\delta ^2}\)
Given:
| Statistical Figures | Value | Origin |
|---|---|---|
| \(z_{\alpha }\) | 1.645 | based on the previous equation |
| \(z_{\beta }\) | 1.29 | since the β should be less than 0.1 |
| \(\sigma\) | 1.25 | given |
| \(\delta\) | 4 | from 44 - 40 |
| \(\mu _0\) | 44 | given mean |
\(n=\frac{\left(1.65 + 1.29\right)^2 1.25 ^2}{4 ^2}\)
n = 0.8441
Rounding this off to the nearest whole number will give us 1 as sample size.
E. Explain how you could answer the question in part A by calculating an appropriate confidence bound on battery life.
In calculating an appropriate confidence bound by upper-tailed test, we will use the formula:
\(\left(-\infty ,\:\bar{x}+z_a\frac{\sigma }{\sqrt{n}}\right)\)
Given:
| Statistical Figures | Value |
|---|---|
| \(z_{\alpha }\) | 1.645 |
| \(\bar{x}\) | 40.5 |
| \(\sigma\) | 1.25 |
| n | 10 |
\(\left(-\infty ,\:40.5+1.645\frac{1.25}{\sqrt{10}}\right)\)
\(\left(-\infty ,\:41.15\right)\)
The mean 40 is within the confidence bound so we fail to reject the null hypothesis.There is no evidence to support the claim that battery life exceeds 40 hours.
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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PARAMETER OF INTEREST:
Difference between the mean of the samples.
\(\bar{x}_2-\bar{x}_1\:and\:\Delta _0\)
NULL HYPOTHESIS:
\(\mu_2-\mu_1\:=\Delta_0\):
ALTERNATIVE HYPOTHESIS:
\(H_1:\:\:\mu_2>\mu_1\)
TEST STATISTIC:
\(T_0=\frac{\bar{x}_2-\bar{x}_1-\Delta _0}{\sqrt{\frac{S_2^2}{n_2}+\frac{S_1^2}{n_1}}}\)
REJECT \(H_0\) IF:
\(v=\frac{\left(\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}\right)^2}{\frac{\left(\frac{s_1^2}{n_1}\right)^2}{n_1-1}+\frac{\left(\frac{s_2^2}{n_2}\right)^2}{n_2-1}}\)
Given:
| Statistical Figures | Value |
|---|---|
| \(S_1\) | 0.4 |
| \(S_2\) | 0.6 |
| \(n_1\) | 16 |
| \(n_2\) | 16 |
\(v=\frac{\left(\frac{0.4^2}{16}+\frac{0.6^2}{16}\right)^2}{\frac{\left(\frac{0.4^2}{16}\right)^2}{16-1}+\frac{\left(\frac{0.6^2}{16}\right)^2}{16-1}}\)
v = 26.13
v = 26
Thus, the critical value will be 1.706.
Reject the null hypothesis if the P-value is less than α (0.05)
Reject the null hypothesis if the 1.706 is less than \(T_0\).
COMPUTATION:
\(T_0=\frac{\bar{x}_2-\bar{x}_1-\Delta _0}{\sqrt{\frac{S_2^2}{n_2}+\frac{S_1^2}{n_1}}}\)
Given:
| Statistical Figures | Value |
|---|---|
| \(S_1\) | 0.4 |
| \(S_2\) | 0.6 |
| \(n_1\) | 16 |
| \(n_2\) | 16 |
| \(\bar{x}_1\) | 20.2 |
| \(\bar{x}_2\) | 19.6 |
| \(\Delta _0\) | 0 |
\(T_0=\frac{20.2-19.6-0}{\sqrt{\frac{0.4^2}{16}+\frac{0.6^2}{16}}}\)
\(T_0\)= 3.328
For the p-value we have:
pt(3.328201177, df=26.13402062, lower.tail=FALSE)## [1] 0.001302688
P= 0.001
CONCLUSION:
CRITICAL NUMBER < \(T_0\)
1.705 < 3.328
We reject the null hypothesis because the critical number is less than the \(T_0\).
P-value < α
0.001 < 0.05
Additionally, the p-value is also less than α (0.05).
Therefore, there are evidences formed from the tests that support the statement that the mileage of B is significantly better when compared to A.
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