document.write( "Question 1206353: In a survey, 13 people were asked how much they spent on their child's last birthday gift. The results were roughly bell-shaped with a mean of $35.6 and standard deviation of $3.3. Estimate how much a typical parent would spend on their child's birthday gift (use a 99% confidence level). Give your answers to 3 decimal places.\r
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Algebra.Com's Answer #843761 by Theo(13342)![]() ![]() You can put this solution on YOUR website! sample size is 13. \n" ); document.write( "sample mean is 35.6 \n" ); document.write( "standard deviation is 3.3\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "tk-score formula is t = (x-m)/s \n" ); document.write( "t is the t-score \n" ); document.write( "x is the critical sample mean \n" ); document.write( "m is the sample mean \n" ); document.write( "s is the standard error\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "at 99% two tail confidence interval with 12 degrees of freedom (sample size minus 1 = degrees of freedom in this problem), the critical t-score is equal to plus or minus t = 3.054540 rounded to 6 decimal places.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "s = standard error = standard deviation / sqrt(sample size) = 3.3 / sqrt(13) = .915255 rounded to 6 decimal places.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "t-score formula becomes plus or minus 3.054540 = (x - 35.6) / .915255.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "on the high side of the confidence interval, solve for x to get x = 3.054540 * .915255 + 35.6 = 38.395683 rounded to 6 decimal places.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "on the low side of the confidence interval, solve for x to get x = -3.054540 * .915255 = 35.5 = 32.804317 rounded to 6 decimal places.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "your solution is that the a typical parent would spend from 32.804 to 38.396 on their child's birthday, rounded to 3 decimal places, at 99% confidence interval.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |