document.write( "Question 1096633: It has been reported that the average credit card debt for college seniors at the college book
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document.write( "store for a specific college is $3262 and the standard deviation is $1100. The student senate at
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document.write( "a large university feels that their seniors have debt much less than this, so it conducts a study of
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document.write( "50 randomly selected seniors and finds that average debt is $2995. Is the student senate correct
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document.write( "with 0.05. \n" );
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Algebra.Com's Answer #711159 by Boreal(15235)![]() ![]() You can put this solution on YOUR website! Assuming a z test here with known sd. \n" ); document.write( "the 95% CI, if this is used will be mean +/- 1.96*1100*sigma/sqrt(n) \n" ); document.write( "this is +/- $304/90 or $305 \n" ); document.write( "The 95% CI is ($2690, $3300). Because $3262 is in the confidence interval, one cannot conclude that this is statistically significant at the 0.05 level \n" ); document.write( "The z-test would be (2995-3262)/1100/sqrt(50)=-1.72 \n" ); document.write( "With Ho being that there is no difference \n" ); document.write( "Ha being there is a difference \n" ); document.write( "alpha being 0.05 \n" ); document.write( "test statistic of z=(bar-mean)/sigma/sqrt(n) \n" ); document.write( "critical value is |z|>1.96 \n" ); document.write( "p-value is 0.086, which is greater than 0.05 so Ho is not rejected. \n" ); document.write( " |