Question

Question 2 of 2 < > -/1 View Policies Current Attempt in Progress The life in hours of a battery is known to be approximately normally distributed, with standard deviation - 1.25 hours. A random sample of 10 batteries has a mean life of x = 40.5 hours. (a) Is there evidence to support the claim that battery life exceeds 40 hours? Use a = 0.045. The battery life significantly different greater than 40 hours at a -0.045. (b) What is the P-value for the test in part (a)? P-value Round your answer to three decimal places (e.g. 98.765). (c) What is the B-error for the test in part (a) if the true mean life is 42 hours? B Round your answer to five decimal places (eg. 98.76543). (d) What sample size would be required to ensure that does not exceed 0.2 if the true mean life is 44 hours? batteries Statistical Tables and Charts e Textbook and Media Sunit Answer Attempts: 0 of 3 used Save for Later

Answer 1

a) The test statistic here is computed as:

$z^* = \frac{\bar X - \mu_0}{\frac{\sigma}{\sqrt{n}}} = \frac{40.5 - 40}{\frac{1.25}{\sqrt{10}}} = 1.2649$

As this is a one tailed test, therefore the p-value here is obtained from standard normal tables as:
p = P(Z > 1.2649) = 0.1030

As the p-value here is 0.1030 > 0.045, which is the level of significance, therefore the test is not significant here and we cannot reject the null hypothesis here. Therefore the battery life is not significant different greater than 40 hours at the given level of significance here.

b) The p-value computed in the previous part here is 0.1030

Therefore 0.1030 is the required p-value here.

c) For 0.045 level of significance, we have from standard normal tables:
P(Z < 1.695) = 0.955 that is P(Z > 1.695) = 1 - 0.955 = 0.045

Therefore the critical mean value here is computed as:

$\bar X_{crit} = \mu_0 + 1.695*\frac{\sigma}{\sqrt{n}} = 40 + 1.695*\frac{1.25}{\sqrt{10}} = 40.67$

Now for a true mean value of 42, the probability of type 2 error is computed as the probability of not rejecting the null hypothesis which is computed here as:

$P(\bar X < 40.67)$

Converting it to a standard normal variable, we have here:

$P(Z < \frac{40.67 - 42}{\frac{1.25}{\sqrt{10}}})$

Getting it from the standard normal tables, we have here:

therefore 0.0004 is the required probability here.

d) For a beta error of 0.2, we have from standard normal tables:
P(Z < -0.842) = 0.2

Therefore the sample size here is computed as:

$\frac{40 + 1.695*\frac{1.25}{\sqrt{n}} - 44}{\frac{1.25}{\sqrt{n}}} = -0.842$

$-3.2\sqrt{n} + 1.695 = -0.842$

Therefore 1 is the required minimum sample size required here.