Can someone help me with PART G ONWARDS Here\'s my file: https://docs.google.com

Question

Can someone help me with PART G ONWARDS

Here's my file:https://docs.google.com/spreadsheets/d/1Bt9ccQVdcENZLq17iHEcWypKJMMI7JWJkl6dSjJzu8U/edit?usp=sharing

g) For each of the four estimators compute the mean of the sample of 3000 (should be close to Zero h) For each of the four estimators compute the root mean square (RMS) about zero. (SUMPRODUCT anyone?) i) For each of the four estimators use Excel's STDEV.S function to compute the standard deviation. How do these values compare with the results of h)?

Explanation / Answer

Answer for part g)

First of arrange the data in ascending order:

the mean for mean and median would be same as:

-0.003299116892

Formula used: =average(E2:E3002)

.

The Mean for truncated mean data would be:

0.2041017057

.Formula used is: =average(A3:A3001)

.

The mean for the double truncated mean data would be:

0.08493311295

.Formula used is: =Average(A3:A3000)

.

Answer to part h)

The formula to be used for RMS is :

=sqrt(sumsq(E2:E3002)/3001)

For mean and median we get: 1.015963775

.

The RMS value for truncated data is :

=sqrt(sumsq(E3:E3001)/2999)

We get : 1.013055524

.

The RMS value for double truncated data is:

=sqrt(sumsq(E4:E3000)/2997)

We get 1.013512378

.

Answer to part i )

For mean and median the value of standard deviation can be calculated with the formula

=stdev.s(E2:E3002)

For truncated data: =stdev.s(E3:E3001)

For double truncated data: =stdev(E4:E3000)

.

Answer to part j)

The Ratios of RMS of mean to other three RMS are as follows:

RMS (mean) / RMS (median) = 1.015963775 / 1.015963775 = 1

RMS (mean) / RMS (truncated) = 1.015963775 / 1.013055524 = 0.9997703145

RMS (mean) / RMS (double truncated) = 1.015963775 / 1.013512378 = 0.9996768726

.

We find the least ratio is for double trucated data, thus it provides the most robust data. when we eliminate the largest and the smallest values and trucate the data, we end up getting data that is more close by , thus reducing the deviations, and hence reducing the error values, thus resulting in highly robust data

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