document.write( "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.
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document.write( "90 of the 1594 randomly selected smart phones broke before the warranty expired.\r
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document.write( "a. With 95% confidence the proportion of all smart phones that break before the warranty expires ______ and ______.\r
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document.write( "b. If many groups of 1594 randomly selected smart phones are selected, then a different confidence interval would be produced for each group.
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document.write( "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. \n" );
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Algebra.Com's Answer #812963 by Boreal(15235) You can put this solution on YOUR website! p hat is 90/1594=0.0565 \n" ); document.write( "half-interval is z(0.975)*sqrt(phat*(1-phat)/n) \n" ); document.write( "=1.96*sqrt(0.0565*0.9435/1594) \n" ); document.write( "=0.0113 \n" ); document.write( "the interval is (0.0452, 0.0678) \n" ); document.write( "- \n" ); document.write( "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. \n" ); document.write( " |