Question

A tablet advertises that it has an all-day battery. The company claims that their battery will expectedly last 24 hours. Batteries of that capacity are assumed to have lives that are normally distributed with (known) standard deviation equal to 1.25 hours. A random sample of 10 batteries has a sample average life of 23.2 hours. Answer the following questions.

(a) Using α = 0.05, can you support the claim that battery life is below 24 hours? Hint: Formulate the problem with H0 : µ = µ0 and H1 : µ < µ0.

(b) What is the P-value for the test in part (a)?

(c) What is the β error associated with the test assuming the true mean of the battery is 22 hours?

(d) What sample size would be required to ensure that β is less than or equal to 0.1 assuming the true mean is 22 hours?

Question 1: Testing a mean A tablet advertises that it has an all-day battery. The company claims that their battery will expectedly last 24 hours. Batteries of that capacity are assumed to have lives that are normally distributed with (known) standard deviation equal to 1.25 hours. A random sample of 10 bat- teries has a sample average life of 23.2 hours. Answer the following questions. (a) Using a 0.05, can you support the claim that battery life is below 24 hours? HINT: Formulate the problem with Ho : /u =140 and H1 : u < po (b) What is the P-value for the test in part (a)? (c) What is the B error associated with the test assuming the true mean of the battery is 22 hours? (d) What sample size would be required to ensure that B is less than or equal to 0.1 assuming the true mean is 22 hours? Answer:

Answer 1

a)

(1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: u = 24 Ha: u < 24 This corresponds to a left-tailed test, for which a z-test for one mean, with known population standard deviation will be used. (2) Rejection Region Based on the information provided, the significance level is a = 0.05, and the critical value for a left- tailed test is ze = -1.64. The rejection region for this left-tailed test is R = {2:2 < -1.64} (3) Test Statistics The z-statistic is computed as follows: X - uo c/VT 23.2 - 24 1.25/10 = -2.024 (4) Decision about the null hypothesis Since it is observed that 2 = -2.024 < ze = -1.64, it is then concluded that the null hypothesis is rejected. Using the P-value approach: The p-value is p=0.0215, and since p = 0.0215 < 0.05, it is concluded that the null hypothesis is rejected. (5) Conclusion It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population mean p is less than 24, at the 0.05 significance level.

Since the null hypothesis is rejected we have evidence to support the claim that  battery life is below 24 hours.

b) The p-value of the test is 0.0215

c) This is a left tailed test.

We will fail to reject the null (commit a Type II error) if we get a Z statistic greater than -1.64.

This -1.64 Z-critical value corresponds to some X critical value such that

Xc – 24 1.25 -= -1.64 V10

X = 23.35

So I will incorrectly fail to reject the null as long as a draw a sample mean that greater than 23.35.

To complete the problem what I now need to do is compute the probability of drawing a sample mean greater than 23.35 given µ = 22. Thus, the probability of a Type II error is given by

P(Z > 23.35 - 22 1.25 TO = P(Z > 3.4153)

P(Z > 3.4153) =1- P(Z <3.4153) = 1-0.9997 = 0.0003

B = 0.0003

d) The formula for sample size is

2(Z + Z3)- n = AR

$\Delta$ is the effect size

B=0.01, a = 0.05

Zo = 20.025 = 1.96

Z3 = 20.01 = 2.3263

= 2(1.96 +2.3263)*(1.25)? - (24-22) 15