Question

A manufacturer of nickel-hydrogen batteries randomly selects 100 nickel plates for test cells, cycles them a specifled number of times, and determines that 11 of the plates have blistered. (a) Does this provide compelling evidence for conduding that more than 10% of all plates blister under such circumstances? State and test the appropriate hypotheses using a significance level o 0.05 @ Ho: ρ-D.10 Ha: ρ > 0.10 Ho: P 0.10 Ha , p z 0.10 0.10 Ha : H0: ρ > 0.10 Calculate the test statistic and determine the p-value. Round your test statistic to two decimal places and your ρ-value to four decimal places.) P-value State the conclusion in the problem context. Reject the null hypothesis. There is sufficient evidence that more than 10% of plates blister under the experimental oonditions. Do not relect the null hypothesis. There is not suffident evidence that more than 10% of plates blister under the experimental oonditions Do not reject the null hypothesis. There is sufficient evidence that more than 10% of plates blister under the experimental conditions. ● Reject the null hypothesis. There is not sufficient evidence that more than 10% of plates blister under the experinental conditions. In reaching your conclusion, what type of error might you have committed? type I ⓔ typer! b) If it is really the case that 1ō% of all plates blister under these c rcumstances and a sample size 100 is used how likely isit that the null hypothesis of part a will not be rejected by the 0.05 test? Round your answer to four decimal places. If it is really the case that 109h af all plates blister under thase arcumstances and sample sze 20D ls used, how likely is it that the null hypothesis of part will not be rejected by tha D 0ฐ test? Round your answer to four decimal places. a (c) How many plates would hava to he tasted to hava 0.16) 0.10 for the test of part (a)? (Round your answer up to the naxt whale number.) plates

Answer 1

Here we have following information:

$n=100, \hat{p}=\frac{11}{100}=0.11$

Hypotheses are:

$H_{0}:p\leq 0.10$

$H_{a}:p>0.10$

Test is right tailed.

Standard deviation of the proportion is:

$\sigma=\sqrt{\frac{p\left ( 1-p \right )}{n}}=\sqrt{\frac{0.10\left ( 1-0.10 \right )}{100}}=0.03$

Test statistics will be:

$z=\frac{\hat{p}-p}{\sigma}=\frac{0.11-0.10}{0.03}=0.33$

Alternative hypothesis shows that the test is right tailed so p-value of the test is

P(z > 0.33) = 0.3707

Since p-value is greater than significance level of 0.05 so we fail to reject the null hypothesis. That is there is no evidence to conclude that more than 10% of all plates blister under such circumstances.

(b)

Here we need to find the type II error. For 0.05 significance level, critical value of z is 1.645. The critical value of sample proportion for which we will reject the null hypothesis is

$z=\frac{\hat{p}-p}{\sigma}$

$1.645=\frac{\hat{p}-0.10}{0.03}$

$\hat{p}=0.14935$

Now the standard deviation for p = 0.16 is

$\sigma=\sqrt{\frac{p\left ( 1-p \right )}{n}}=\sqrt{\frac{0.16\left ( 1-0.16 \right )}{100}}=0.0367$

The z-value such that $\hat{p}=0.14935$ and p = 0.16 is

$z=\frac{0.14935-0.16}{0.0367}=-0.29$

The required probability is

$\beta=P(z<-0.29)=0.3859$

c)

Standard deviation of the proportion is:

$\sigma=\sqrt{\frac{p\left ( 1-p \right )}{n}}=\sqrt{\frac{0.10\left ( 1-0.10 \right )}{200}}=0.0212$

The critical value of sample proportion for which we will reject the null hypothesis is

$z=\frac{\hat{p}-p}{\sigma}$

$1.645=\frac{\hat{p}-0.10}{0.0212}$

$\hat{p}=0.134874$

Now the standard deviation for p = 0.16 is

$\sigma=\sqrt{\frac{p\left ( 1-p \right )}{n}}=\sqrt{\frac{0.16\left ( 1-0.16 \right )}{200}}=0.0259$

The z-value such that $\hat{p}=0.134874$ and p = 0.16 is

$z=\frac{0.134874-0.16}{0.0259}=-0.97$

The required probability is

$\beta=P(z<-0.97)=0.166$

d)

Here test is right tailed. And we have

$p_{1}=0.16,p_{0}=0.10$

The effect size is

$ES=\frac{p_{1}-p_{0}}{\sqrt{p_{0}(1-p_{0})}}=\frac{0.16-0.10}{\sqrt{0.10(1-0.10)}}=0.2$

For $\alpha=0.05$ we have $z_{1-\alpha}=1.645$

and for 90% power we have $z_{1-\beta}=1.28$

So the requried sample size is

$n=\left ( \frac{z_{1-\alpha}+z_{1-\beta}}{ES} \right )^{2}=\left ( \frac{1.645+1.28}{0.2} \right )^{2}=213.890625$

Hence, required sample size is 214.