Given Data:
Sample size \(n = 10\)
Sample mean \(\bar{x} = 40.5\)
Population standard deviation \(\sigma = 1.25\)
Significance Level \(\alpha = 0.05\)
Applying the 7-step procedure for Hypothesis Testing,
1. Parameter of Interest: The parameter of interest is the mean battery life \(\mu\).
2. Null hypothesis: \(H_0 : \mu = 40\)
3. Alternative hypothesis: \(H_1 : \mu > 40\)
The problem calls for an upper-tailed test.
4. Test Statistic:
Since the population standard deviation is known, the test statistic is solved using the equation:
\[\begin{aligned} Z_0 = \frac{\bar{x} - \mu_0}{\frac{\sigma}{\sqrt{n}}} \end{aligned}\]5. Reject \(H_0\) if: Reject \(H_0\) if the \(P\)-value is less than \(\alpha = 0.05\).
6. Computations
Substituting the values:
\[\begin{aligned} Z_0 = \frac{40.5 - 40}{\frac{1.25}{\sqrt{10}}} = 1.264911 \end{aligned}\]We can compare the resulting \(P\)-value of the test statistic with the significance level \(\alpha\). Because the test is an upper-tailed test, we need to compute for the probability that a random variable is greater than the test statistic \(Z_0\)
Since the \(P\)-value or \(P_v\) is the smallest significance level that can lead to the rejection of the null hypothesis, if \(\alpha = 0.05\) is less than \(P_v\), there is not enough evidence to reject the null hypothesis.
pnorm(1.2649111, lower.tail = F)## [1] 0.10295167. Conclusion: Since \(P_v\) has a value of 0.1029516, which is greater than \(\alpha\), there is not enough evidence to reject the null hypothesis, and in turn, support the claim that the battery life exceeds \(40\) hours.
The \(P_v\) is equal to \(0.1029516\).
The \(\beta\)-error, for this question, is the probability that the null hypothesis is not rejected even though it is false.
For an upper-tailed test is given by the solution:
\[\begin{aligned} \beta = P(Z_\alpha - \frac{\delta\sqrt{n}}{\sigma}) \end{aligned}\]where \(\delta = \mu - \mu_0, \delta > 1\)
The values we input are the following:
We can get \(Z_\alpha\) by getting the z-score of our significance level \(\alpha = 0.05\) from the right.
qnorm(0.05, lower.tail = F)## [1] 1.644854\(Z_\alpha = 1.644854\)
\(\bar{X} = 42\)
\(\mu = 40\)
\(n = 10\)
\(\sigma = 1.25\)
Substituting the values:
\[\begin{aligned} \beta = P(1.644854 - \frac{(42 - 40)\sqrt{10}}{1.25}) = P(-3.41479) \end{aligned}\] \[\begin{aligned} \beta = 0.0003191557 \end{aligned}\]The formula for the sample size for a one-sided test is given as:
\[\begin{aligned} n = \frac{(Z_\alpha + Z_\beta)^2\sigma^2}{\delta^2} \end{aligned}\]\(Z_\alpha =\) 1.6448536
\(Z_\beta =\) 1.2815516
\(\delta = 4\)
Substituting the values:
\[\begin{aligned} n = \frac{(1.644854 + 1.281552)^21.25^2}{4^2} = 0.8363137 \approx 1 \end{aligned}\]A sample size of \(1\) is required to ensure that the \(\beta\)-error does not exceed \(0.10\) if the true mean is \(44\).
Since the test is an upper-tailed test, we only need to find the upper endpoint of the confidence interval.
This is given by the interval \((-\infty, \bar{X} + \frac{Z_\alpha \sigma}{\sqrt{n}})\).
\(Z_{0.95} =\) 1.6448536
\[\begin{aligned} 40.5 + \frac{(1.644854)(1.25) }{\sqrt{10}} = 41.15019 \end{aligned}\]At an upper-tailed test with \(\alpha = 0.05\), we can say with \(95\%\) confidence that the true mean is in the interval \(\mu \leq 41.15019\). Since \(40\) is in the interval, there is not enough evidence to reject the null hypothesis \(H_0: \mu = 40\), and therefore, not enough evidence to support the claim that the battery life exceeds \(40\) hours
References:
D. C. Montgomery and G. C. Runger, in Applied Statistics and Probability for Engineers, 7th ed., Hoboken, NJ: Wiley, 2018, pp. 193–258.