A sample is analyzed in triplicate, giving the results below. The method used fo

Question

A sample is analyzed in triplicate, giving the results below. The method used for the analysis has been performed thousands of times before, giving a standard deviation of 88.0. Calculate the 99% confidence interval using the format average(±range). Your answers must be within 1 of the exact answers to be counted correct so don't do too much rounding. Report your answers to the nearest 0.1

7635,   7497,   7496

Confidence level (%) Degrees of freedom 99.9 50 1.000 6.3142.706 31.821 63.656 127.321 636.578 0.816 2.920 4.303 6.965 9.92514.089 31.598 0.7652.353 3.182 4.5415.841 0.7412.132 2.776 3.7474.6045.598 0.727 2.0152.571 3.3654.032 4.773 0.718 1.943 2.447 3.143 3.7074.3175.959 0.711 1.895 2.365 2.998 3.500 4.029 5.408 0.7061.860 2.306 2.896 3.355 0.7031.833 2.262 2.821 3.2503.690 4.781 0.700 1.812 2.228 2.764 3.1693.581 0.6911.753 2.131 2.602 2.9473.252 4.073 0.6871.725 2.086 2.528 2.845 0.6841.708 2.060 2.485 2.787 0.6836972.042 2.457 2.7503.030 3.646 0.6811.684 2.021 2.423 2.704 0.679.671 2.000 2.390 2.660 2.9153.460 0.677 1.658 1.980 2.358 2.617 2.860 3.373 0.674645 1.960 2.326 2.576 2.8073.291 90 95 98 99.5 2 7.4532.924 8.610 6.869 4 3.832 5.041 4.587 3.850 10 15 20 25 30 40 60 120 3.153 3.078 3.725 2.971 3.551

Explanation / Answer

Standard error of the sampling distribution of mean = Std Deviation / sqrt(n)

= 88 / sqrt(1000) = 2.783

Mean of the triplicate = (7635 + 7497 + 7496) / 3 = 7542.667

The degree of freedom = Number of samples - 1 = 3 -1 = 2

Significance level = 1 - Confidence level = 1 - 0.99 = 0.01

For a two tail, the signficance level = 0.01/2 = 0.005

Confidence level for two tail = 1 - 0.005 = 0.995 = 99.5%

From the table, for 99.5% confidence level and degree of freedom 2,

Test statistic = 14.089

So, margin of error = Test statistic * Standard error

= 14.089 * 2.783 = 39.20969

So, The true value is determined to be 7542.67 ( ± 39.21 ) at the 99% level of confidence

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