We Have Numbers Of Free Samples

Introduction:

Truly the requirement of some statistical analysis is very much important for some kind of data we have for any organization or any country. What we have done in this report is very much necessary to evaluate with some few words as a gist in this part. As an introduction part we can say the report consists of very basic two topics one is about the data of master degree students in a particular organization and another is about the house prices of inside London and out side of London. On these two very basic topics we have performed very detail analysis in terms of statistical parameters. Now the question may arise, what are the analyses we have done on this report? To make this question clear we have to say all the detail analysis we have done in this report. To elaborate every thing in a short paragraph is not enough. So we have discussed everything with the analysis files and calculations. For both of the two cases we have done everything those we need for related problem statement we have in our hand. To prove them and to clear everything on such a manner so that everyone can understand we have done all the software automatic calculations as well as mathematical calculation manually.

TASK: 1

1. The first thing is to find out the population mean and the variance of the population ages for the problem statement given here,

Now calculation is being done on the basis of functions and formula available in excel we have found out the mean and the variance of the population mean.

Mean of the population age is 22.607 and the variance of the population age is 2.232784. It has been calculated by using the format “=AVERAGE(A1:A1000) and =STDEVA(A1:A1000)” then by squaring standard deviation we got the variance.

Average	22.607
Standard deviation	1.49425
Variance	2.232784

 

2. To determine the 30 sample values having 100 numbers of population on each sample of the population data given we have prepared a list of the 30 random sample and having 100 elements randomly taken from the 1000 numbers data sheet.

From that data sheet we obtained after finding out the 30 sample data sheet we just need to find out two different estimator means according to the problem statement.

a. First thing we need to draw the 30 samples of 100 number of population which we have done using random sampling function in Microsoft excel.

The table provided consists of 30 rows signifies 30 samples and having 100 rows signifies 100 number of population taken as a sample paramente randomly from the data sheet.

 

Along with this a snap shot is also given as an evidence to check whether we have the sampling correctly or not.

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a. For this segment we need to calculate the mean for each 30 sample data list having 100 numbers of element that is ¯x₁ and for the same sampling data we need to fig out a another estimator mean also by taking the first 20 sample from each sample list among the 30.

For each set of sampled data we performed the function “=AVERAGE(A1:A100)” and got the result as ¯x₂ after that we have made a new workbook by taking first 20 values from the 100 number of values for each sample set and got the averages of them with the same manner.

The mathematical formula is kind of this,

For the first case,

For the second case,

For the output result we have find out the results and make a table containing the results, as found out is excel analysis.

This table signifies the values of means for the two process first column is for total 100 numbers mean for 30 sample and second column signifies the mean of 30 sample for first 20 numbers ages in the list.

b.

¯x1	¯x2
23	23.3
22	22.4
23	23
23	22.9
22	22.2
23	22
23	23
23	23
23	23
23	23.1
23	22.4
22	23
23	22
23	23
23	22.5
22	23
23	22.4
22	22.3
23	23
23	22.7
22	22
23	23
23	22
23	22
22	23
23	23
23	23
22	23
22	23
23	23

 

This snap shot signifies that we have made a new workbook of first 20 parameters taken from 30 samples and calculated the mean for each sample.

1. For finding out the bias for both the cases for and it is very easy to calculate. We just need to calculate the mean of the estimations of first case then if we subtract the reference value from the mean then we will get the bias value.

For first case (¯x₁) the value of mean is 22.58883797 and for the second case () the mean is 22.64923977. Now as a reference value we have taken 23 for the first case and for the second case the reference value is considered also 23.

Therefore, the bias value for the first case is;

Bias for first estimatior is, 22.58883797 – 23 = -0.4112 (corrected upto four decimal places)

Bias for second estimator is,

22.64923977 – 23 = -0.3508 (corrected upto four decimal places)

Yes, the bias for the second case is higher than that of the second case.

The reason behind higher bias in second case due to the less number of samples and one more reason is that the mean value for the second case is higher than that of the second case.

2. Here in this part we need to calculate the variance for the both cases in first estimator and in the second estimator.

Calculating variance is not so tough; it is just the squared value of the standard deviation. Now we need to findout the standard deviation for the two cases and then we can find out the variance of them.

To calculate the variance for the two individual estimators we have used function to formulate the standard deviation “=STDEV (initial: ending)” for both the cases. As a result, we have got the variance 0.023 for the first case and 0.095 for the second case. Here the variance for the first case is smaller than that of the variance of the second case. That means we can say easily the first one or estimator 1 is most efficient. The reason behind it is, the number of samples taken into consideration is larger than the number of samples taken in second case. So the output value for the first will be much more efficient is very obvious.

TASK 2:

1. Here we are calculating the confidence intervals for 99% of confidence value. For both the value of mean for inside London and mean for outside London we are doing the same thing. Since we have not the accessibility of the STATA software tools, we are using the Microsoft Excel and mathematical calculations for this calculation.

a. For the initial case we are finding out the confidence interval for the mean inside London,

The mathematical formula for finding out confidence interval is,

For 99% of confidence interval we can say the upper bound is,

And for the lower bound it is,

As a whole we can say the confidence intervals at 99% of confidence is,

First we need to calculate the value (average of the price inside the London). That is, 146.4484.

Then the requirement is to calculate the standard deviation for the prices inside London. That is 48.47424.

Now we are going to calculate the sample size which is n. the value of n is 171.

The confidence coefficient for 99% confidence is 2.576.

Now the margin of error is. That has a value of 9.549021

Now for fin ding out the upper bound we just need to add the mean with the margin of error,

= 146.4484 + 9.549021
= 155.9975

Just like the same method we need to calculate the lower bound for the mean price outside London,

= 146.4484 – 9.549021
= 136.8994

Table: we get this table for 99% confidence interval for mean inside London from Excel.

Average	146.4484
Standard deviation	48.47424
Sample size	171
Confidence coefficient	2.576
Margin of error	9.549021
Upper bound	155.9974
Lower bound	136.8994
Max	245
Min	50
Range	195

 

b. Now for this case we are finding out the confidence interval for the mean outside London,

First we need to calculate the value ¯x_{outsidelondon} (average of the price outside the London). That is, 95.90687).

Then the requirement is to calculate the standard deviation for the prices outside London. That is 44.35201

Now we are going to calculate the sample size which is n. the value of n is 329

The confidence coefficient for 99% confidence is 2.576.

Now the margin of error is . That has a value of 6.298849

Now for fin ding out the upper bound we just need to add the mean with the margin of error,

= 95.90687 – 6.298849
= 89.60802

Table: we get this table for 99% confidence interval for mean outside London from Excel.

Average	95.90687
Standard deviation	44.35201
Sample size	329
Confidence coefficient	2.576
Margin of error	6.298849
Upper bound	102.2057
Lower bound	89.60802
Max	236
Min	11
Range	225

 

As instructed we have done this whole calculation in excel by applying the mathematical formula for finding out the confidence intervals by considering the mean price inside London and outside London. First we have separated the entire excel file into two segments like inside London sheet and outside London sheet. Then by applying the functions available in excel we have done this whole thing.

As a result, we have prepared the list of observation from this entire analysis,

For 99% confidence interval the intervals are 224.2492 to 68.64764 (for inside London) and the intervals are 102.2057 to 89.60802 (for outside London).

2. For consideration we have got the sample file with it, but according to the formula we have in our hand we have practically calculated the confidence intervals for both the cases where we considered both the lists for finding out the same for inside London as well as outside London.

After doing this all we have come to know that the average values are there in the sample file much higher the reason behind it is, we have considered the excel file provided to us for data collection. But the data founded out by our calculation are completely based on the sample file and the formula given there. For our analysis we have come to know that the result we have got interval 224.2492, 68.64764 for 99% confidence interval in case of mean price inside London. And the interval 102.2057, 89.60802 for 99% confidence interval in case of mean price outside London.

The data we have found out from our physical analysis and calculation is perfectly matched with the sample file calculations, only difference is the outcome values. Since we have different values in the list, we got different mean as well as different intervals at 99% confidence level.

3. Here we just need to calculate the number of samples required if the mean in case of houses to be purchased in London having a mean of GBP 5000. If we calculate the value of n with considering the upper bound for 99% confidence interval, then we will get a negative value of n. So we have to calculate the value of n with considering the lower bound value of 99% confidence interval value. Then only we will get a positive value for number of sample.

It is given that,

At the time when we apply the value 5000 on average it will automatically show a lower bound of 4990.451 and the upper bound is 5009.549 for 99% confidence interval. Applying these values on equation we will get the same value of sample number of 171.

From this equation we have to calculate the value of n required.

This is the basic samples required for making the mean GBP 5000 for the house price in London. This only for making a proof of concept practically it is not possible to make a number of sample a fractional number.

4. Here we need to calculate the 99% confidence interval for the differences in the mean prices of properties in London and properties outside London.

As the variance is unknown here and it is said that the variance is same for both the cases, that is why are using the pooled sample variance tom calculate this.

For our case the number of samples for inside London is 171 and for outside of London it is 329.

Finding out pooled sample variance,

The value of pooled sample variance is 1156345984

Now for calculating 99% confidence interval for difference in mean prices, the formula is

That means the interval on 99% confidence for the difference of mean prices in London and outside of London is kind of this -3155.2324<< 3256.315

5. We have calculated the 99% confidence interval for the difference in the mean price of the houses inside London and outside London. Since we have calculated the variance from the two individual excel sheet from their standard deviation values. And as the assignment says the variance fir both cases are same, that is why we used the concept of pooled sample variance and get to know the exact interval for the two different cases mean values.

The values of confidence interval at 99% confidence level are 3256.315 or -3115.2324 for the data sheen we have in our hand.

6.

a. Now, for this part the problem statement is average families will not be able to afford a house in London since the mean house price is increased to more than GBP150000.

This statement was published on a news paper in that time. We need to do the justification if it is reasonable to grant this statement or it is not to be grant.

For doing this thing we have to perform mathematical calculation or analysis in excel.

Regression statistics

Multiple R	0.265606813
R square	0.070546979
Adjusted R square	0.068680608
Standard error	49.86152267
Observation	500

 

	df	SS	MS	F	Significance F
Regression	1	93974.79671	93974.8	37.799	1.60963E-09
Residual	498	1238113.379	2486.171
Total	499	1332088.175

 

	Coefficients	Standard error	T stat	P value	Lower 95%	Upper 95%	Lower 95%	Upper 95%
Intercept	135.9011429	4.314549811	31.49812	1.3E-120	127.4241201	144.378166	127.42412	144.378166
X variable	13.37400526	2.175310866	-6.14809	1.61E-09	-17.64792314	-9.1000874	-17.6479321	9.10008739

 

From the analysis we can get conclusion that, the claim of that news paper is not valid of the analysis. Since not a single price was more than GBP150000. That is completely against the data sheet. Since the data sheet gives us a mean for the house price in London is about 146.4484 so there is not any chance like no one affords a house in London since the mean is more than GBP150000. This is the statement signifies the report results.

b. For testing this set of data in 5% of significance level, it means the hypothesis is to be taken or it can be rejected. For this purpose, our approach is to find out the p value in terms of statistics. And the comparison is required to the significance level. As the significance figure is 5% or can be said it is 0.05

Here the basic fundamental rule is terms of rejecting any hypothesis considering its significance level is kind of that,

If the p value is less than the value of 0.05 for this case then we can reject the hypothesis, and if the p value is greater or equals to 0.05 then we cannot reject the hypothesis.

For the p value theorem here it is 1.61E-09 which is very less than 0.05 so we can easily reject the hypothesis stated here.

c. As the result interpretation we have come to know that, the regression analysis is the most important analysis method to find out the statistical analysis throughout. From the regression analysis we have got so many things related to the data sheet we have in our hand. From here we get to know the value of R, value of R square and also the p values and the s values of the data sheet we have. The p value plays an important role for rejecting or not rejecting any hypothesis statement just we have seen in the previous points. These are the basic results and their useful interpretation on the context of data analyzing may be it is done on software tool or it is done mathematically does not matters.

7. It is basically the conclusion part of this whole experimental analysis we have till now in this report. Very basic things we have done in this report like the mean declaration for individual two cases house prices for inside London and for outside London. Not only the mean, along with the mean we have calculated the standard deviation, variance and margin of error. These are not the only outcomes. Along with them we have also done so many complex calculations for finding out the confidence intervals at 99% confidence level. For both the cases we have found out the lower bound and the upper bound (intervals). From that interval we will get to know the upper estimated possible value and lower possible estimated value. We have observed all the formulas and by using them we have prepared the mathematical calculations. Secondly we have calculated the confidence intervals for mean difference for two cases like inside London and outside London.

Considering the last case that is the news paper report says about the mean price that is more than GBP 150000. On considering this value, we have done detail regression analysis to find out the rejection criteria for the value stated on news paper. We have approached the p value theorem to conclude the rejection criteria of that statement on 5% of the significance level. From this data analysis we get to know there is a pure chance to reject the statement given in the news paper.

Conclusion:

As we have done with the whole statistical analysis for the ages of master degree students for the task one and another statistical analysis on house prices in side of London and outside of London. We have gathered so many knowledge on very complex statistical analysis. The automatic sum and calculation methodology in Microsoft excel. The findings of task 2 are performed on mathematical calculation without using STATA. So for this we have done so many calculations on statistics. The first task is complete done on Microsoft excel and an appropriate result is in our hand. Some of the snapshots are also given in this report so that one can understand clearly, what have done actually to do this calculation and how it is performed?

What are the criteria for rejecting any statistical statement by using hypothesis analysis is also in this report? With the help of finding out the variance we have also concluded the most efficient estimation to be taken and the reason behind it. Each and every detail decisions is there for the two tasks and their problem statements.

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