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Parameter of interest: \(\mu\), the mean battery life
Null Hypothesis,\(H_0\):
\[ H_0: \mu = 40 \]
\[ H_1: \mu > 40\]
\[ z_0 = \frac{\bar{x} - \mu_0}{\sigma \sqrt{n}} \]
\[ Reject~H_0~if~z_0 > z_{0.05},~z_{0.05}=~ 1.65 \]
\[ z_0 = \frac{40.5 - 40}{1.25 \sqrt{10}} = 1.26 \]
Using z0 = 1.26 and using the formula for P in an upper tailed test: \[ P ~=~ 1-\Phi(z_0) \]
P-Value: \[ P = 1 - \Phi(z_0) ~=~ 1 - \Phi(1.26) ~=~1 - 0.896165 = 0.103835 \] Hence the P-value of the test is P = 0.103835.
Here, we let \(\mu\) = 42 given the condition that the true mean life of the battery is 42 hours
To find the \(\beta\)-error, we use the formula for \(\beta\) in a one tailed test(upper bound): \[ \beta = \Phi(z_\alpha - \frac{\delta \sqrt{n}}{\sigma}),~where~\delta = \mu - \mu_0 \]
Using z\(\alpha\) = 1.65, \(\delta\) = \(\mu\) - \(\mu_0\) = 42 - 40 = 2, \(\sigma\) = 1.25. We get
\[ \beta ~= ~\Phi(1.65 - \frac{2\sqrt{10}}{1.25})~= ~\Phi(-3.41) ~= ~0.000325 \]
So the \(\beta\)-error when true mean life is 42 hours is \(\beta\) = 0.000325
This time, we let \(\mu\) = 44. We need to find n so that \(\beta\) < 0.10 when the true mean life of the battery is 44 hours using the formula:
\[ n = \frac{(z_\alpha + z_\beta)^2 \sigma^2}{\delta^2}, ~where ~\delta = \mu - \mu_0 \]
So far we have z\(\alpha\) = 1.65 and \(\sigma\) = 1.25
Since we let \(\mu\) = 44 and use \(\mu\)0 = 40, we can get \(\delta\) = \(\mu\) - \(\mu\)0.
\(\delta\) = \(\mu\) - \(\mu\)0 = 44 - 40 = 4
For z\(\beta\), since \(\beta\) < 0.10, we let z\(\beta\) = \(z_{1-0.1}\) = z0.9 = 1.28
Substituting the values we get
\[ n = \frac{(1.65 + 1.28)^2(1.25)^2}{4^2} = 0.838369 = 1 \] So n = 1. This means that in order to ensure that \(\beta\) does not exceed 0.10 when the true mean life is 44 hours, the sample size must be 1.
Going back to part A,
Parameter of interest: \(\mu\), the mean battery life
Null Hypothesis,\(H_0\): \[ H_0: \mu = 40 \]
Alternative Hypothesis, \(H_1\): \[ H_1: \mu > 40\]
The condition states that we calculate using confidence bound or interval wherein \(\alpha\) = 0.05. the confidence interval on \(\mu\) is given by 100(1-\(\alpha\)). So the confidence interval on \(\mu\) is \[ 100(1-0.05)\% = ~95\% \]
So, we are finding the 95\(\%\) coverage for \(\mu\), the mean battery life.
Since this is a one-tailed test, the 95\(\%\) coverage is 50\(\%\) below the mean, which is denoted by: \[ (-\infty, \bar{x} ~+ ~z_\alpha \frac{\sigma}{\sqrt{n}}) \]
So we use:
\[ \bar{x} ~+ ~z_\alpha \frac{\sigma}{\sqrt{n}} \]
So for us to reject \(H_0\)
\[ Reject~H_0~if~~\mu_0 > \bar{x} ~+ ~z_\alpha \frac{\sigma}{\sqrt{n}} \]
because we want to find out if the mean battery life exceeds 40 hours.
Given \(\mu_0\) = 40 (for comparison on the confidence interval),
\(\bar{x}\) = 40.5, \(\sigma\) = 1.25, n = 16, and \(z_\alpha\) = z0.05 = 1.65 (since \(\alpha\) = 0.05)
\[ 40.5 ~+ ~1.65\frac{1.25}{\sqrt{16}} ~= ~41.0156 \] So the one-tailed test confidence interval of 95% on \(\mu\) is \[ (-\infty, 41.0156) \] Conclusion:
Since \(\mu\)0 = 40 is inside the confidence interval or \(\mu\)0 less than 41.0156 \[ -\infty < ~\mu_0~=~40 ~< ~41.0156 \] We fail to reject H0 and conclude that there is no evidence that supports the claim that the mean battery life exceeds 40 hours.
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From the question above, we are given:
| Brand A | Brand B |
|---|---|
| \(\bar{x}\)1 = 19.6 | \(\bar{x}\)2 = 20.2 |
| S1 = 0.4 | S2 = 0.6 |
| n = 16 | n = 16 |
With the variances unknown, we must choose between:
Case 1: \(\sigma_1^2\) = \(\sigma_2^2\) = \(\sigma^2\) wherein we assume that the variances are equal. If this case is used, we will use a pooled estimator of \(\sigma^2\) defined by the equation:
\[ S_p^2~= ~\frac{(n_1-1)S_1^2~+~(n_2-1)S_2^2}{n_1 ~+~ n_2 ~-~ 2} \]
and given the assumptions, the quantity/statistic is defined by: \[ t_0 ~=~ \frac{\bar{x}_1 -\bar{x}_2 - \Delta_0}{S_p\sqrt{\frac{1}{n_1}~+~\frac{1}{n_2}}} ~~~~~with ~degrees ~of ~ freedom ~= n_1 +n_2-2 \]
or
Case 2: \(\sigma_1^2\) \(\neq\) \(\sigma_2^2\) wherein we assume that the variances are unequal.If this case is used, the statistic is:
\[ t_0 ~=~ \frac{\bar{x}_1-\bar{x}_2-\Delta_0}{\sqrt{\frac{S_1^2}{n_1}+\frac{S_2^2}{n_2}}} \] distributed approximately as t with degrees of freedom given by: \[ v ~=~ \frac{(\frac{S_1^2}{n_1}+\frac{S_2^2}{n_2})^2}{\frac{(S_1^2/n_1)^2}{n_1-1}+\frac{(S_2^2/n_2)^2}{n_2-1}} \]
To determine which case to use, We look at the sample standard deviations S1 = 0.4 and S2 = 0.6
The sample standard deviations are similar to each other so I think we can assume that the variances are equal or \(\sigma_1^2\) = \(\sigma_2^2\) = \(\sigma^2\) so we use case 1.
Our parameter of interest is \(\mu_1\) and \(\mu_2\), the mean mileage of the gasoline brands.
Null Hypothesis, \(H_0\): \(H_0\): \(\mu_1\) - \(\mu_2\) = 0 or \(\mu_1\) = \(\mu_2\).
Alternative Hypothesis, \(H_1\): \(H_1\): \(\mu_1\) < \(\mu_2\).
Test Statistic:
\[ t_0 ~=~ \frac{\bar{x}_1 -\bar{x}_2 - \Delta_0}{S_p\sqrt{\frac{1}{n_1}~+~\frac{1}{n_2}}} ~~~~~with ~degrees ~of ~ freedom ~= n_1 +n_2-2 \]
Reject H0 if:
Using the P-value (for a one-tailed test):
\[ Reject ~H_0 ~if ~P{-value} ~<~0.05 \]
Using the fixed significance level \(\alpha\) = 0.05 and degrees of freedom = 16 + 16 - 2 = 30 (for a one-tailed test),
\[ Reject ~H_0 ~if ~t_0<-t_{\alpha,n_1+n_2-2}= ~-t_{0.05,30} ~= ~-1.697 \]
Computations:
Using the given data and the pooled estimator equation, the pooled estimated deviation Sp is:
\[ S_p^2 ~=~ \frac{(16-1)(0.4)^2~+~(16-1)(0.6)^2}{n_1+n_2-2}~=~ 0.26 \] \[ S_p ~=~ \sqrt{0.26} ~=~ 0.509901 ~=~ 0.510 \] Substituting all the values including Sp to the test statistic equation and \(\Delta\)0 = 0, \[ t_0 ~=~ \frac{19.6-20.2-0}{0.510\sqrt{\frac{1}{16}+\frac{1}{16}}} ~=~ -3.32756 ~=~ -3.328 \]
So, our \(t_0\) lies between \(t_{0.001,30}\) = 3.385 and \(t_{0.0025,30}\) = 3.030 or 3.030 < 3.328 < 3.385.
In line with this, the P-value of \(t_0\) is (0.001 < P(\(t_0\)) < 0.0025).
Note: We can also assume that the P-value of \(t_0\) is close to 0.001 since \(t_0\) is much closer to \(t_{0.001,30}\) = 3.385.
Hence, Using P-value approach and \(\alpha\) = 0.05: \[ (0.001 < P(-3.328) < 0.0025) < 0.05 ~ or ~P(t_0)~=~P(-3.328)<0.05 \] Using the fixed significance level approach: \[ t_0<-t_{\alpha,n_1+n_2-2}~~= t_0 < -t_{0.05,30}~=~-3.328 < -1.697 \] Since P(-3.328) < 0.05 and \(t_0\) = -3.328 < \(t_{0.05,30}\) = -1.697, we reject \(H_0\).
Conclusion:
Since we reject \(H_0\), we can say that there is strong evidence that the mileage of brand B is significantly greater than brand A.