Question 1

The life in hours of a battery is known to be approximately distributed with standard deviation \(\sigma\) = 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 \(\alpha\) = 0.05.

For this problem, we can answer it using the seven-step procedure recommended for hypothesis-testing [1].

Step 1. Parameter of Interest

The parameter of interest for this problem is the \(\mu\), or the mean battery life.

Step 2. Null Hypothesis

\(H_0\): \(\mu\) = 40 hours

Step 3. Alternative Hypothesis

\(H_1\): \(\mu\) > 40 hours

We will be formulating our hypothesis using one-sided alternative hypothesis since we will be testing if there is an evidence that supports the claim that battery life exceeds 40 hours. From this, the claim is that the battery life is exceeding 40.

\[ \begin{aligned} H_0~: \mu = 40 \\ H_1~: \mu > 40 \\ \end{aligned} \]

Step 4. Test Statistic

The test statistic formula is given as: \[ Z_0 = \frac{\bar{x} - \mu}{\sigma / \sqrt {n}} \]

Step 5. Reject \(H_0\) if:

Reject \(H_0\) if the P-value is less than 0.05. To use a fixed significance level test, the boundaries of the critical region would be \(Z_.05 = 1.64\).

Step 6. Computations

From the given data:

\[ \begin{aligned} n = 10 \\ \sigma = 1.25 \\ \bar{x} = 40.5 \\ \alpha = 0.05 \\ \end{aligned} \]

We will be having a rough sketch of the bell curve for us to know the region for rejection and region for failed to reject.

95% confidence level

Therefore, the \(Z_\alpha\) or the z for critical value will be 1.64 for the confidence level of 95% since \(\alpha\) is 0.05 or 5%. Remember this because we need to establish our decision rule that if \(z > 1.65\), we must reject \(H_0\). From the picture above, the blue shaded region is the rejection region and the the rest of the graph is failed to reject region.

We will be using the z-test since the sample size is less than 30, so the normal distribution will work for this problem. Now, we solve for the test statistic, \(Z\), given the formula: \[ Z_0 = \frac{\bar{x} - \mu}{\sigma / \sqrt {n}} \]

Substituting the data given from the problem, we now have: \[ Z_0 = \frac{40.5 - 40}{1.25/ \sqrt {10}} \]

And this will be equal to: \[ Z_0 = 1.26 \]

By using the P-value method we shall arrive at the following conclusions: \[ \begin{aligned} p-value < \alpha \text {(We must reject} H_0 \text{.)}\\ p-value \ge \alpha \text{(We fail to reject} H_0 \text{.)}\\ \end{aligned} \]

Since the Z-value is 1.26 from the computation earlier, we will be looking at the positive z-score table and found out that the correspondent area to the left is 0.8962. Here is the positive z-score table for reference.

Positive Z-score Table

Now, we have to subtract 0.8962 from 1 to calculate the area on the right of the curve. This will give us 0.1038. This is the area of the blue shaded region on the right.

Now that we have the P-value, which is 0.1038, we need to compare it to \(\alpha\) which is 0.05.

\[ 0.1038 > 0.05 \]

Step 7. Conclusions

Since the P-value is greater than the alpha, we fail to reject the null hypothesis, \(H_0\): \(\mu\) = 40. This is to say that there is no enough evidence to support the claim that the battery life exceeds 40 hours.

Counter checking this with the graph, we know that 1.26 (the z-value) is less than 1.64, or in the region wherein we fail to reject the null hypothesis, this means that there is no enough evidence to support the claim that the battery life exceeds 40 hours.

Whatever method we will use is that we will arrive at the same conclusion.

B. What is the P-value for the test in Part A?

To solve for the P-value for the test in Part A, we first found out that the Z-value is 1.26 by using the test statistic formula. Upon looking at the positive z-score table, we knew that the correspondent area to the left of the curve is 0.8962. We then proceed in subtracting this from 1, to get the corresponding area to the right of the curve which will give us the value of 0.1038. The p-value is the smallest level of significance that would lead to rejection of the null hypothesis \(H_0\) with the given data [2].

The computation is as follows: \[ \begin{align} P-value &= 1-\phi(Z_0)\\ &= 1-\phi(1.26)\\ &=1-0.8962\\ &=0.1038 \end{align} \]

Notice that we got the same P-value for this problem that was used in part A. Therefore, the P-value for the test 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?

A Type II Error is defined as failing to reject the null hypothesis when it is false [3]. For this problem, we are asked to compute for the probability of having a type II error when the true mean life is 42 hours.

From Part A, we have the following hypothesis: \[ \begin{aligned} H_0~: \mu = 40 \\ H_1~: \mu > 40 \\ \end{aligned} \] and the given that follows: \[ \begin{aligned} n = 10 \\ \sigma = 1.25 \\ \bar{x} = 40.5 \\ \alpha = 0.05 \\ \end{aligned} \] From the sketch below, we know that the rejection region is the blue shaded region with 0.05 or 5% which is the alpha and the rest of the curve, 95%, is the acceptance region.

95% confidence level

The picture above shows the standard normal distribution wherein the \(\mu\) is transformed into zero and the same rejection region of 0.05 or 5%. Now, referring to the positive z-scores table above, we found that the corresponding z-score of 0.05 is 1.65 that tells us about the area of the region to the right of the cut-off of the rejection region.

To compute for the critical value, c, we have this formula: \[ c = {\mu} + Z_c ( \frac{\sigma} {\sqrt {n}}) \] Plugging in the known values, we now have: \[ c = {40} + 1.64 ( \frac{1.25} {\sqrt {10}}) \] And we will have: \[ c = 40.65 \] This means that the cut-off for the rejection region is at 40.65, which is at the right of the mean. Now, we knew that 42 is true mean as stated in the problem, the critical value, 40.65 is at the left of the true mean value. Because we are finding the probability of having a type II error, we will find the area of the region to the left of 40.65. Now, we have to plug it in the formula for test statistic which is: \[ Z_0 = \frac{\bar{x} - \mu}{\sigma / \sqrt {n}} \] And substituting the known values will give us, \[ Z_0 = \frac{40.65-42}{1.25 / \sqrt {10}} \] That will give us \(Z_0 = -3.42\). If we refer to the negative z-scores table below, we know that its corresponding z-value is 0.0003.

Negative z-scores Table

Another way to solve it is to use the formula for \(\beta\) that is: \[ \beta = \phi(Z_\alpha- \frac{\delta \sqrt{n}}{\sigma}) \] We have the following values: \[ \begin{aligned} n &= 10 \\ \sigma &= 1.25 \\ \delta &= \mu-\mu_0 \implies \delta = 42-40 = 2 \\ Z_\alpha &= 1.64 \end{aligned} \] Substituting the known values will give us: \[ \beta = \phi(1.64- \frac{2 \sqrt{10}}{1.25}) \] The result will be \[ \begin{aligned} \beta = \phi(-3.42) \\ \beta = 0.0003 \end{aligned} \]

Notice that we got the same answer. Therefore, the \(\beta\)-error if the true mean life is 42 hours will be 0.0003 or 0.03%.

D. What sample size would be required to ensure that \(\beta\) does not exceed 0.10 if the true mean is 44 hours?

For this problem, we will be using the formula for sample size for a one-sided test on the mean with variance being known which is: \[ n \simeq \frac{(Z_\alpha +Z_\beta)^{2} \sigma^{2}}{\delta^{2}} \] wherein, \(\delta =\mu-\mu_0\).

We know the following values: \[ \begin{aligned} \sigma = 1.25 \\ Z_\alpha=1.64 \\ \delta=44-40 \implies \sigma=4 \\ \end{aligned} \] To find \(Z_\beta\), since the \(\beta\) should not exceed 0.10, \(\beta<0.10\), and we will find for \(Z_.10\). In accordance to the positive z-score table, it is known that \(Z_\beta=1.28\).

We now plug in these values to the formula and we now have: \[ \begin{aligned} n &\simeq \frac{(1.64 +1.28)^{2} (1.25)^{2}}{(4)^{2}}\\ n &\simeq \frac{(8.5264) (1.5625)}{16}\\ n &\simeq \frac{13.3225}{16}\\ n &\simeq 0.83265625\\ n &\simeq 1\\ \end{aligned} \]

Thus, the sample size required to ensure that \(\beta\) does not exceed 0.10 if the true mean is 44 hours will be 1.

E. Explain how could you answer the question in part A by calculating an appropriate confidence bound in battery life.

In this problem, we are dealing with a normal distribution that has a population that we do not have any idea about its value. However, since the problem gave us the confidence level of 95% (this is from the value of \(\alpha=0.05\)), we can use this along with the following given from the data.

\[ \begin{aligned} n = 10 \\ \sigma = 1.25 \\ \bar{x} = 40.5 \\ \end{aligned} \] From the given information above, we know that \(\mu\) is between \(\bar{x}- (Z_\alpha)(\sigma / \sqrt{n})\).

\[\bar{x}- (Z_\alpha)(\sigma / \sqrt{n}) \leq \mu \leq \bar{x} + (Z_\alpha)(\sigma / \sqrt{n})\]

The confidence interval is going to be the value from \(\bar{x}\) minus the error bound for the mean and \(\bar{x}\) plus the error bound for the mean. \[CI = (\bar{x} - EBM, \bar{x} + EBM)\]

The error bound for the mean (EBM) is the \((Z_\alpha)(\sigma / \sqrt{n})\).

95% confidence level

The area that corresponds to the left of the \(Z_\alpha\) is the region shaded in light blue that can help us calculate the area to the of \(Z_\alpha\) that is 1.64. Now, we have this formula for the confidence interval: \[CI=(\bar{x}- (Z_\alpha)(\sigma / \sqrt{n}) , \bar{x} + (Z_\alpha)(\sigma / \sqrt{n}))\]

Now that we have the known values, we just have to substitute it to the formula and we now have: \[ \begin{aligned} CI &=(\bar{x}- (Z_\alpha)(\sigma / \sqrt{n}) , \bar{x} + (Z_\alpha)(\sigma / \sqrt{n}))\\ CI &=(40.5- (1.64)(1.25 / \sqrt{10}) , 40.5 + (1.64)(1.25 / \sqrt{10}))\\ CI &=(39.8517 , 41.1483)\\ \end{aligned} \]

We got the confidence interval which is CI = (39.8517 , 41.1483) hours meaning that all the values in the interval are considered as plausible values for the parameter being estimated. If the value of the parameter being specified by the null hypothesis, which in our case is \(H_0 = 40\), is present in the 95% confidence interval, then the null hypothesis cannot be rejected at 0.05 level. On the other hand, if the value of the parameter being specified by the null hypothesis is not present in the interval, then the null hypothesis can be rejected at 0.05 level [4].

Now, we can formulate our conclusion based on the conditions above. Notice that the value of the parameter in the null hypothesis we have, \(H_0 = 40\), is included in the interval (39.8517 , 41.1483). This means that the null hypothesis cannot be rejected at 0.05 level or fail to reject the null hypothesis at 0.05 level. Notice that we have the same conclusion in part A. This is to say that there is no enough evidence to support the claim that the battery life exceeds 40 hours..

Therefore, we can calculate for the appropriate confidence bound or the confidence interval to test the hypothesis but we need to be careful in determining the interval to arrive at the correct conslusion.

Question 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 \(\alpha\) = 0.05 approaches (if possible).

The first thing we have to do is to list the values given from the data.

For Brand A: \[ \begin{aligned} n_1 &= 16 \\ \sigma_1 &= 0.4 \\ \bar{x}_1 &= 19.6 \\ \end{aligned} \] For Brand B: \[ \begin{aligned} n_2 &= 16 \\ \sigma_2 &= 0.6 \\ \bar{x}_2 &= 20.2 \\ \end{aligned} \] This problem has small samples taken, wherein both are less than 40, there is an assumption that the populations are normally distributed and we will be basing our test for hypotheses and confidence intervals on the t distribution. Also, in this problem, we will be dealing it using case number 2, test statistic for the difference in means wherein variances are unknown and assumed unequal [5].

For Case 2, if \(H_0: \mu_1 - \mu_2 = \Delta_0\) is true, the test statistic with formula, \[T^*_0=\frac{\bar{x}_1-\bar{x}_2-\Delta_0}{\sqrt{\frac{S^2_1}{n_1}+\frac{S^2_2}{n_2}}}\]

is distributed approximately as t with degrees of freedom given by \[v=\frac{(\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2})^2}{\frac{(\frac{s^2_1}{n_1})^2}{n_1-1}+\frac{(\frac{s^2_2}{n_2})^2}{n_2-1}}\]

If v is not integer, round down to the nearest integer.

For this problem, we will be using the seven-step hypothesis-testing process.

Step 1. Parameter of Interest

The parameters of interest are \(\mu_1\) and \(\mu_2\), and we want to know if \(\mu_1 - \mu_2 = 0\).

Step 2.Null Hypothesis

The null hypothesis is \(H_0: \mu_1 - \mu_2 = 0\) or \(H_0: \mu_1 = \mu_2\).

Step 3. Alternative Hypothesis

The alternative hypothesis is \(H_0: \mu_1 \neq \mu_2\)

Step 4. Test Statistic

The test statistic is given by this formula: \[t^*_0=\frac{\bar{x}_1-\bar{x}_2-0}{\sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}}\]

Step 5. Reject \(H_0\) if:

The degrees of freedom is given by the formula above. We need to compute for the degrees of freedom to find the crtitical t-values. Therefore,

\[ \begin{aligned} v &=\frac{(\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2})^2}{\frac{(\frac{s^2_1}{n_1})^2}{n_1-1}+\frac{(\frac{s^2_2}{n_2})^2}{n_2-1}}\\ v &=\frac{[\frac{(0.4)^2}{16}+\frac{(0.6)^2}{16}]^2}{\frac{[\frac{(0.4)^2}{16}]^2}{16-1}+\frac{[\frac{(0.6)^2}{16}]^2}{16-1}}\\ v &=\frac{[\frac{1}{100}+\frac{9}{400}]^2}{\frac{[\frac{1}{100}]^2}{15}+\frac{[\frac{9}{400}]^2}{15}}\\ v &=\frac{\frac{169}{160,000}}{\frac{[\frac{1}{100}]^2}{15}+\frac{[\frac{9}{400}]^2}{15}}\\ v &=\frac{\frac{169}{160,000}}{\frac{97}{2,400,000}}\\ v &=26.134\\ v &\approx26 \end{aligned} \] The degrees of freedom is 26.

Therefore, using \(\alpha\) = 0.05 and a fixed significance-level test, we will reject \(H_0: \mu_1 = \mu_2\) if \(t^*_0 > t_{0.025,26}=2.056\) or if \(t^*_0 < t_{0.025,26}=-2.056\).

The table of t-critical values is attached below for reference.

Table of t-critical values

Step 6. Computations

The given from the data are as follows:

For Brand A: \[ \begin{aligned} n_1 &= 16 \\ \sigma_1 &= 0.4 \\ \bar{x}_1 &= 19.6 \\ \end{aligned} \] For Brand B: \[ \begin{aligned} n_2 &= 16 \\ \sigma_2 &= 0.6 \\ \bar{x}_2 &= 20.2 \\ \end{aligned} \]

We now substitute the known values to the test statistic formula we will be using: \[ \begin{aligned} t^*_0&=\frac{\bar{x}_1-\bar{x}_2-0}{\sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}}\\ t^*_0&=\frac{19.6-20.2}{\sqrt{\frac{(0.4)^2}{16}+\frac{(0.6)^2}{16}}}\\ t^*_0&=\frac{-0.6}{\sqrt{\frac{1}{100}+\frac{9}{400}}}\\ t^*_0&=\frac{-0.6}{\sqrt{\frac{13}{400}}}\\ t^*_0&\approx-3.33\\ \end{aligned} \]

The test statistic, \(t^*_0\), is -3.33.

Now, we will use the p-value method and see if we can come up with the same conslusion later on.

For the p-value method, we have the following conclusions: \[ \begin{aligned} p-value < \alpha \text {(We must reject} H_0 \text{.)}\\ p-value \ge \alpha \text{(We fail to reject} H_0 \text{.)}\\ \end{aligned} \] The p-value is 0.001303 for a t-score of -3.33 with degrees of freedom of 26. Looking at the relationship between the p-value and the significance level, \(\alpha\), 0.001303 < 0.05.

Step 7. Conclusions

By using fixed significance level test, since \(t^*_0=-3.33 < t_{0.025,26}=-2.056\), we reject the null hypothesis.

By using p-value method,0.001303 < 0.05, we conclude that we reject the null hypothesis.

Decision

Since we have rejected the null hypothesis that Brands A and B are the same, we want to know now if Brand B is better in mileage than in Brand A. The p-value (0.001303) being less than the level of significance (0.05) means there is no enough evidence that can support the null hypothesis that claims Brand A has the same performance or better than Brand B. Thus, Brand B has a greater mileage than Brand A.

References:

[1]D. C. Montgomery and G. C. Runger, Applied Statistics and Probability for Engineers. 2018.

[2]D. C. Montgomery and G. C. Runger, Applied Statistics and Probability for Engineers. 2018.

[3]D. C. Montgomery and G. C. Runger, Applied Statistics and Probability for Engineers. 2018.

[4]“Interpreting confidence levels and confidence intervals (article),” Khan Academy. https://www.khanacademy.org/math/ap-statistics/estimating-confidence-ap/introduction-confidence-intervals/a/interpreting-confidence-levels-and-confidence-intervals (accessed Jul. 26, 2021).

[5]D. C. Montgomery and G. C. Runger, Applied Statistics and Probability for Engineers. 2018.