Statistics Marathon & Questions (1 Viewer)

Trebla · Jun 2, 2016

University Statistics Discussion Marathon

Expand the square and note that the mean (and also its square) is a constant.

In the second summand there is a common factor which can be factorised out.

In the third summand note that you are summing the same constant value n times.

davidgoes4wce · Jun 2, 2016

Re: University Statistics Discussion Marathon

davidgoes4wce said:
$Consider a sample size n=10 where the first sample mean is 8 and sample variance is 13. Consider a second sample that is exactly the same as the first excerpt that there is an additional observation that takes on the value of 8. What is the sample variance for this larger sample with n=11 observations? (Your answer should be correct to 1 decimal place.)$

Ans : 11.7

$I managed to figure it out$

$s^2=\frac{1}{n-1} \sum ^{10} _{i=1} (y_{i}-\bar{y})^2=\frac{1}{n-1} (\sum_{i=1} ^{10} y_i ^2 -n \bar{y}^2)$

$s^2=13, \bar{y}=8, n=10 . $ We can then $$

$13=\frac{1}{9} (\sum_{i=1} ^{10} y_i ^2 -10 (8)^2)$

$\sum_{i=1} ^{10} y_i ^2 =757$

$$ With n=11 our new summand is , $ \sum_{i=1} ^{11} y_i ^2 =757+8^2=821, n=11, \bar{y}=8$

$s^2=\frac{1}{n-1} (\sum_{i=1} ^{10} y_i ^2 -n \bar{y}^2)$

$=\frac{1}{10} (821-11(8)^2)=11.7$

seanieg89 · Jun 2, 2016

Re: University Statistics Discussion Marathon

Here is something more theoretical.

Suppose $X_j\sim \mathcal{N}(\mu,\sigma^2)$ for j=1,...,n are i.i.d random variables for some unknown parameters $\mu,\sigma^2$ .

1. Define $\overline{X}:=\frac{1}{n}\sum_{j=1}^n X_j, s^2:=\frac{1}{n-1}\sum_{j=1}^n (X_j-\overline{X})^2.$

Compute $\mathbb{E}(\overline{X}),\mathbb{E}(s^2), \textrm{Var}(\overline{X}) , \textrm{Var}(s^2).$

2. Hence define in terms of $\overline{X},s,n$ a random variable that has expected value $\mu$ and variance $1$ . Compute the pdf of this random variable in terms of special functions.

3. What happens as $n\rightarrow\infty$ ? Prove this.

(This question outlines some of the theory behind something used several times in this thread.)

BlueGas · Jun 2, 2016

Re: University Statistics Discussion Marathon

Can you guys stop dropping the big guns... you're making statistics look hard.

seanieg89 · Jun 2, 2016

Re: University Statistics Discussion Marathon

BlueGas said:
Can you guys stop dropping the big guns... you're making statistics look hard.

Huh? Not all of statistics is just plugging numbers into memorised formulae, where do you think these formulae come from?

As always, you have the option of ignoring any question not to your taste.

In any case, this particular question is easier than most of the mathematical questions posted in these forums.

davidgoes4wce · Jun 3, 2016

Re: University Statistics Discussion Marathon

seanieg89 said:
Here is something more theoretical.

Suppose $X_j\sim \mathcal{N}(\mu,\sigma^2)$ for j=1,...,n are i.i.d random variables for some unknown parameters $\mu,\sigma^2$ .

1. Define $\overline{X}:=\frac{1}{n}\sum_{j=1}^n X_j, s^2:=\frac{1}{n-1}\sum_{j=1}^n (X_j-\overline{X})^2.$

Compute $\mathbb{E}(\overline{X}),\mathbb{E}(s^2), \textrm{Var}(\overline{X}) , \textrm{Var}(s^2).$

2. Hence define in terms of $\overline{X},s,n$ a random variable that has expected value $\mu$ and variance $1$ . Compute the pdf of this random variable in terms of special functions.

3. What happens as $n\rightarrow\infty$ ? Prove this.

(This question outlines some of the theory behind something used several times in this thread.)

$I'll have a crack at Question 1$

$Defining $ \bar{X} $ and s^2$

$$ If the n observations in a sample are denoted by $ \bar{x}=\frac{x_1+x_2+....+x_n}{n}=\frac{\sum _{i=1}^n x_i}{n} \textcircled{1}$

$Computation of s^2.$ $ The computation of $ s^2 $ requires calculation of $ \bar{x}, $ n subtractions, and n squaring and adding operations. If the original observations or the deviations $ x_i-\bar(x} $ are not integers, the deviations $ x_i-\bar{x} $ may be tedious to work with, and several decimals may have to be carried to ensure numerical accuracy. A more efficient computational formula for the sample variance is obtained as follows:$

$s^2=\frac{\sum_{i=1}^n (x_i-\bar{x})^2}{n-1} =\frac{\sum_{i=1}^n (x_{i}^2+\bar{x}^2-2\bar{x}x_{i})}{n-1}$

$and since $ \bar{x}=\frac{1}{n} \sum_{i=1}^n x_i, $ the last equation reduces to$

$s^2=\frac{\sum_{i=1}^n x_{i}^2-\frac{(\sum_{i=1}^n x_i)^2}{n}}{n-1}=s^2=\frac{\sum_{i=1}^n (x_i-\bar{x})^2}{n-1}$

$$ Taking the expectation of both sides in $ \textcircled{1}$

$E[\bar{X}]=\frac{1}{n} \sum_{i=1}^nE[X] \textcircled{2}$

$Since $ X_1,X_2,....,X_n $ are independent and identically distributed (i.i.d.) RVs , we have E[X_1]=E[X_2] $ =...E[X_n]=\mu_x. $ Substitituing these values into $ \textcircled{2}, $ we find that the sample mean variable satisfies$

$E[\bar{X}]=\mu_x$

$Which asserts that $ \bar{x} $ is an unbiased estimate of $ \mu_x. $ An unbiased estimate is one that is, on the average, right on target.$

$Var[\bar{X}]=E[(\bar{X}-E[{\bar{X})]^2]$

$where $ \bar{X}-E[{\bar{X}]=\bar{X}-\mu_x $ can be rewritten as$

$\bar{X}-E[\bar{X}]=\frac{1}{n} \sum_{i=1} ^n (X_i-\mu_x)=\frac{1}{n} \sum_{i=1}^n Y_i$

$Where$

$Y_i \triangleq \ X_i-\mu_x, i=1,2,....,n$

$\therefore, Var[\bar{X}]=E[(\frac{1}{n} \sum_{i=1}^n Y_i)^2]= \frac{1}{n^2} \sum_{i=1}^n E[Y_i ^2] +\frac{1}{n^2} \sum_{i=1}^n \sum_{j=1 \ (j \neq i)}^n E[Y_i Y_j]$

$Since the random variables $ {Y_i; 1 \leq i \leq n} $ are statistically independent with zero mean and variance $ \sigma_x ^2 $ we have$

$Var[\bar{X}]=\frac{\sigma_x ^2}{n}$

$Thus the variance of the sample mean variable is the population variance divided by the sample size.$

$The deviations of the individual observations from the sample mean provide information about the dispersion of the $ x_i $ about $ \bar{x}. $ We define the sample variance $ s_x^2 $ by$

$S_x^2 \triangleq \frac{1}{n-1} \sum_{i=1} ^n (x_i-\bar{x})^2 \textcircled{3}$

$The quantity can be viewed as an instance of the sample variance variable, which is also commonly called the sample variance. We find, after some rearrangement$

$S_x^2=\frac{1}{n} \sum_{i=1}^n Y_i ^2-\frac{1}{n(n-1)} \sum_{i=1}^n \sum_{j=1 (j \neq i)} ^n Y_i Y_j$

$Taking expectations, we have$

$E[S_x ^2]=\frac{1}{n} \sum_{i=1}^n E[Y_i ^2]=\sigma_x ^2$

$The reason for using n-1 rather than n as a divisor in $ \textcircled{3} $ is to make $ E[S_x^2] $ equal to $ \sigma_x^2; $ that is , to make $ s^2 $ an unbiased estimate of $ \sigma_x^2. $ The positive square root of the sample variance,$ s_x,$ is called the sample standard deviation$

seanieg89 · Jun 4, 2016

Re: University Statistics Discussion Marathon

Yep good stuff, it remains to compute Var(s^2), but what you have done is enough to motivate the later parts of the question so don't worry about that if you don't want to.

edwardjoh2 · Jun 4, 2016

Re: University Statistics Discussion Marathon

View attachment Screen Shot 2016-06-04 at 11.14.13 pm.zip

Can someone explain for me please

Paradoxica · Jun 5, 2016

Re: University Statistics Discussion Marathon

edwardjoh2 said:
View attachment 33238

Can someone explain for me please

That's dodgy. Please leave the screenshot in the form of an actual image. I don't want to risk getting an infection.

davidgoes4wce · Jun 5, 2016

Re: University Statistics Discussion Marathon

$A regression analysis was performed to determine which, if any, of protein (grams), fiber (grams), carbohydrates (grams), and sugars (grams) are associated with calories in a sample of 34 breakfast cereals. Which of the following is the null hypothesis for the F-test in this problem?$

$Choose the correct answer below$

$A) All of protein, total fat, fiber, carbohydrates, or sugars help to explain calories in breakfast cereals.$

$B) At least one of protein, total fat, fiber, carbohydrates, or sugars helps to explain calories in breakfast cereals.$

$C) None of protein, total fat, fiber, carbohydrates, or sugars helps to explain calories in breakfast certeals.$

$$ D) Exactly one of protein, total fat, fiber, carbohydrates, or sugars helps to explain calories in breakfast cereals.$

$E) \beta_i=0$

davidgoes4wce · Jun 5, 2016

Re: University Statistics Discussion Marathon

Im thinking the answer is E ( I don't have the solution on me ) Here is my explanation and thinking.

$When there is more than one independent variable, we need a method to test the overall utility of the model. The technique is a version of the analysis .$

$To test the utility of the regression model, we specify the following hypotheses:$

$H_0:\beta_1=\beta_2=....=\beta_k$

$H_a: $ At least one $ \beta_i $ is not equal to zero $$

$IF the null hypotheses is true , none of the independent variables $ x_1,x_2,...,x_k $ is linearly related to y, and therefore the model is useless.$

davidgoes4wce · Jun 5, 2016

Re: University Statistics Discussion Marathon

$If all the data points fall on the least-squares regression line in simple linear regression which of the following is true?$

$A) SS(Regression Model)=SS(Trial)$

$B) F-Statistic=0$

$C) MS(Regression Model)=100 \%$

$D) SS(Total)=0$

$E) None of the above statements are true.$

Rhinoz8142 · Jun 5, 2016

Re: University Statistics Discussion Marathon

A sample of 50 observation is taken from a normal population, with mean of 100 and Standard Deviation 10. If the population is finite with N=250

Find

P(xbar > 103)

davidgoes4wce · Jun 5, 2016

Re: University Statistics Discussion Marathon

Rhinoz8142 said:
A sample of 50 observation is taken from a normal population, with mean of 100 and Standard Deviation 10. If the population is finite with N=250

Find

P(xbar > 103)

$As the population is normally distributed, $ \bar{x} $ is normally distributed. Therefore, as the population standard deviation $ \sigma $ is also known, the standardised test statistic Z has a standard normal distribution.$

$Test statistic :$

$Z=\frac{\bar{X}-\mu}{\sigma \div \sqrt{n}}$

$Z=\frac{103-100}{10/ \sqrt{50}}=2.12$

$P(Z > 2.12) =0.017$

Trebla · Jun 5, 2016

Re: University Statistics Discussion Marathon

Rhinoz8142 said:
A sample of 50 observation is taken from a normal population, with mean of 100 and Standard Deviation 10. If the population is finite with N=250

Find

P(xbar > 103)

$$Since $X\sim N(100,100)$ then $\bar{X}\sim N\left(100,\frac{100}{50}\right)$ so $\\\\P(\bar{X} > 103)\\\\=P\left(\dfrac{\bar{X}-100}{\sqrt{\frac{100}{50}}}>\dfrac{103-100}{\sqrt{\frac{100}{50}}}\right)\\\\=P\left(Z> \dfrac{3}{\sqrt{2}}\right)\:$ where $Z\sim N(0,1)$

then compute the value from there. Note that the population size is not relevant here as we are referring to the distribution of the sample mean.

Rhinoz8142 · Jun 5, 2016

Re: University Statistics Discussion Marathon

davidgoes4wce said:
$As the population is normally distributed, $ \bar{x} $ is normally distributed. Therefore, as the population standard deviation $ \sigma $ is also known, the standardised test statistic Z has a standard normal distribution.$

$Test statistic :$

$Z=\frac{\bar{X}-\mu}{\sigma \div \sqrt{n}}$

$Z=\frac{103-100}{10/ \sqrt{50}}=2.12$

$P(Z > 2.12) =0.017$

dont you have to use the FCF ?

davidgoes4wce · Jun 5, 2016

Re: University Statistics Discussion Marathon

Rhinoz8142 said:
dont you have to use the FCF ?

Think me and Trebla got the same answer, you have to use the n value which in this case is the n=50. (not the finite value)

Rhinoz8142 · Jun 5, 2016

Re: University Statistics Discussion Marathon

davidgoes4wce said:
Think me and Trebla got the same answer, you have to use the n value which in this case is the n=50. (not the finite value)

but don't you use the finite correction factor since n/N = 50/250 = > 0.05. Thus using the equation sqrt(N-n/N-1).

I could also be wrong.

davidgoes4wce · Jun 5, 2016

Re: University Statistics Discussion Marathon

Rhinoz8142 said:
but don't you use the finite correction factor since n/N = 50/250 = > 0.05. Thus using the equation sqrt(N-n/N-1).

I could also be wrong.

if it isn't a 1st year level of statistics then you may be right.

$Finite Population Correction Some formulas used to compute standard errors are based on the idea that (1) samples are selected from an infinite population or (2) samples are selected with replacement. In most actual surveys, neither of these ideas are correct.$

$This does not present much of a problem when the sample size (n) is small relative to the population size (N); that is, when the sample is less than 5% of the population. However, when the sample is larger, it is best to apply a correction to the formulas used to compute standard error (SE). This correction is called the finite population correction or fpc.$

$Here is the formula for the finite population correction, when the random variable is a mean score or proportion.$

$f_{pc} = \sqrt{\frac{N - n}{N - 1}}$

$Here is how the finite population correction is used to compute the standard error of a mean score.$

$SE = [ \frac{standard \ deviation}{\sqrt(n)} ] * f_{pc}$

$And here is how the it is used to compute the standard error of a proportion (p).$

$SE =\sqrt{\frac{p(1 - p)}{n}} * f_{pc}$

$Note that when n is very small relative to N, the finite population correction is almost equal to one; and it has only a small effect on the standard error. However, when n is large relative to N, the finite population correction can have a big effect on the standard error.$

Trebla · Jun 5, 2016

Re: University Statistics Discussion Marathon

Rhinoz8142 said:
but don't you use the finite correction factor since n/N = 50/250 = > 0.05. Thus using the equation sqrt(N-n/N-1).

I could also be wrong.

This is needed if there is a sampling without replacement which I guess now looking at the question again is somewhat implied but not immediately obvious. If it is a sampling with replacement then what I had earlier is correct.

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