SOLUTION: A smart phone manufacturer is interested in constructing a 95% confidence interval for the proportion of smart phones that break before the warranty expires.
90 of the 1594 random
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90 of the 1594 random
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Question 1182850: A smart phone manufacturer is interested in constructing a 95% confidence interval for the proportion of smart phones that break before the warranty expires.
90 of the 1594 randomly selected smart phones broke before the warranty expired.
a. With 95% confidence the proportion of all smart phones that break before the warranty expires ______ and ______.
b. If many groups of 1594 randomly selected smart phones are selected, then a different confidence interval would be produced for each group.
About ________ percent of these confidence intervals will contain the true population proportion of all smart phones that break before the warranty expires and about _______ percent will not contain the true population proportion. Answer by Boreal(15235) (Show Source):
You can put this solution on YOUR website! p hat is 90/1594=0.0565
half-interval is z(0.975)*sqrt(phat*(1-phat)/n)
=1.96*sqrt(0.0565*0.9435/1594)
=0.0113
the interval is (0.0452, 0.0678)
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About 95% of the CI would contain the true pop. proportion and about 5% would not. One would not know which 95 and 5, however.
