A manufacturer of nickel-hydrogen batteries randomly selects 100 nickel plates f
ID: 3312175 • Letter: A
Question
A manufacturer of nickel-hydrogen batteries randomly selects 100 nickel plates for test cells, cycles them a specified number of times, and determines that 11 of the plates have blistered.
(a) Does this provide compelling evidence for concluding that more than 10% of all plates blister under such circumstances?
State and test the appropriate hypotheses using a significance level of 0.05.
Calculate the test statistic and determine the P-value. (Round your test statistic to two decimal places and your P-value to four decimal places.)
(b) If it is really the case that 16% of all plates blister under these circumstances and a sample size 100 is used, how likely is it that the null hypothesis of part (a) will not be rejected by the 0.05 test? (Round your answer to four decimal places.)
c) If it is really the case that 16% of all plates blister under these circumstances and a sample size 200 is used, how likely is it that the null hypothesis of part (a) will not be rejected by the 0.05 test? (Round your answer to four decimal places.)
(d) How many plates would have to be tested to have (0.16) = 0.10 for the test of part (a)? (Round your answer up to the next whole number.)
Explanation / Answer
pcap = 11/100 = 0.11
n = 100
a)
H0: p = 0.1
H1: p > 0.1
SE = sqrt(0.1 * 0.9/100) = 0.03
Test statistics, z = (0.11 - 0.1)/0.03 = 0.3333
p-value = 0.6306
As p-value is greater than the significance level of 0.05, fail to reject null hypo.
b)
c)
d) Less than 10 plates to be tested
p0 (hypothesised proportion) 0.1 SE = sqrt(p*(1-p)/n) 0.03 n 100 alpha 0.05 sample/true proportion 0.16 Std. Error. SE 0.0300 Zcritical 1.645 Xcritical 0.15 Beta or type II error is the probability of fail to reject the null hypothesis P(p>0.15) 0.6388 Hence type II error probability is 0.6388Related Questions
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