Statistics Marathon & Questions (3 Viewers)

BlueGas · May 23, 2016

Re: University Statistics Discussion Marathon

Nice, thanks mate!!

BlueGas · May 25, 2016

Re: University Statistics Discussion Marathon

Need help with this

BlueGas · May 25, 2016

Re: University Statistics Discussion Marathon

BlueGas said:
Need help with this

Bump.

davidgoes4wce · May 26, 2016

Re: University Statistics Discussion Marathon

$\ To test the null hypothesis that the $ u_{1}<u_{2}, $ the procedure is as follows:$

$\ 1. Calculate the difference $ (d_{i}=y_{i}-x_{i}) $ between the two observations on each pair, making sure you distinguish between positive and negative differences$

$$ \ 2. Calculate the mean difference $ \bar{d}$

$$ \ 3. Calculate the standard deviation of the differences, $ s_{d}, $ and use this to calculate the standard error of the mean difference, $ SE(\bar{d})=\frac{s_d}{\sqrt{n}}$

$\ 4. Calculate the t-statistics, which is given by $ T=\frac{\bar{d}}{SE(d)}.$ Under the null hypothesis, this t-statistic follows a t-distribution with n-1 degrees of freedom$

$\ 5. Use a t-table or Excel of the t-distribution to compare your value of T to the $ t_{n-1} $ distribution. This will give the p-value for the paired t-test.$

BlueGas · May 26, 2016

Re: University Statistics Discussion Marathon

davidgoes4wce said:
$\ To test the null hypothesis that the $ u_{1}<u_{2}, $ the procedure is as follows:$

$\ 1. Calculate the difference $ (d_{i}=y_{i}-x_{i}) $ between the two observations on each pair, making sure you distinguish between positive and negative differences$

$$ \ 2. Calculate the mean difference $ \bar{d}$

$$ \ 3. Calculate the standard deviation of the differences, $ s_{d}, $ and use this to calculate the standard error of the mean difference, $ SE(\bar{d})=\frac{s_d}{\sqrt{n}}$

$\ 4. Calculate the t-statistics, which is given by $ T=\frac{\bar{d}}{SE(d)}.$ Under the null hypothesis, this t-statistic follows a t-distribution with n-1 degrees of freedom$

$\ 5. Use a t-table or Excel of the t-distribution to compare your value of T to the $ t_{n-1} $ distribution. This will give the p-value for the paired t-test.$

Thanks for the method, much appreciated!

BlueGas · May 26, 2016

Re: University Statistics Discussion Marathon

The p-value is actually wrong, what degrees of freedom did you use?

davidgoes4wce · May 26, 2016

Re: University Statistics Discussion Marathon

My bad I forgot the degrees of Freedom. For this question it should be 7 degrees of freedom. (My previous working I used a value of 8)

Here is my cell working to get the P-Value on Excel. The P-Value should be 0.052 to 3 decimal places.

BlueGas · May 28, 2016

Re: University Statistics Discussion Marathon

Am I correct for this?

davidgoes4wce · May 28, 2016

Re: University Statistics Discussion Marathon

$$ (a) From the given information . The number of users who are 23 to 35 years old and update their status less often \framebox{74}$

$The expected number of users who are 23 to 35 years old and update their status less often is ,$

$E= \frac{Row \ total \times Column \ total}{Grand \ Total}$

$E= \frac{364 \times 314}{947}=120.7$

$(b) The null and alternative hypothesis are,$

$H_{0}: $ The age is independent of the frequency of the status updates $$

$H_{1}: $ The age is not independent of the frequency of the status updates $$

$The test statistic is 210.9$

$Number of rows r=5$

$Number of columns c=4$

$$ Degrees of Freedom is , DF= (r-1)(c-1)=(5-1)(4-1)=12$

$Using excel function the P-Value is ,$

$$ P-Value=CHIDIST(210.9,12)$

$=0.0000$

$From this P-Value which is low, we reject the Null Hypothesis. Therefore we conclude that the age is not independent of the frequency of the status.$

BlueGas · May 28, 2016

Re: University Statistics Discussion Marathon

I'm getting 1.82215E-38 when using the command =CHIDIST(210.9,12), how did you get 0?

davidgoes4wce · May 28, 2016

Re: University Statistics Discussion Marathon

yeah i rounded my answer to 0.000 , I also got that value of 1.822 E -38

BlueGas · May 28, 2016

Re: University Statistics Discussion Marathon

davidgoes4wce said:
yeah i rounded my answer to 0.000 , I also got that value of 1.822 E -38

So there's no evidence of a relationship?

InteGrand · May 28, 2016

Re: University Statistics Discussion Marathon

Note that for most practical purposes, something on the order of 10^-38 like the Excel calculation there is as good as 0 (since 10^-38 = 0.0000…01, where the 1 comes in the 38th decimal place, so there's thirty-seven 0's after the decimal point).

The E in the Excel thing just means "times 10 to the power of".

BlueGas · May 28, 2016

Re: University Statistics Discussion Marathon

davidgoes4wce said:
$From this P-Value which is low, we reject the Null Hypothesis. Therefore we conclude that the age is not independent of the frequency of the status.$

This is confusing me, if we reject the null hypothesis, shouldn't that mean there is a relationship? Because the null hypothesis assumes there's is no relationship...

BlueGas · May 28, 2016

Re: University Statistics Discussion Marathon

It seems that the word 'independent' confused me lol.

davidgoes4wce · May 28, 2016

Re: University Statistics Discussion Marathon

$We conclude that there is evidence of a relationship between Age and Frequency of Status updates. This means that the age is not independent of the frequency of the updates.$

$The purpose of a Chi-Square Test is to test for Indpendence$

$$ For a test of independence, the null and alternative hypothesis follow:$

$H_{0}:$ The Categorical Variables are Independent (i.e there is no relationship between them) $$

$H_{1}:$ The Categorical Variables are Dependent (i.e there is a relationship between them ) $$

BlueGas · May 28, 2016

Re: University Statistics Discussion Marathon

davidgoes4wce said:
$We conclude that there is evidence of a relationship between Age and Frequency of Status updates. This means that the age is not independent of the frequency of the updates.$

$The purpose of a Chi-Square Test is to test for Indpendence$

$$ For a test of independence, the null and alternative hypothesis follow:$

$H_{0}:$ The Categorical Variables are Independent (i.e there is no relationship between them) $$

$H_{1}:$ The Categorical Variables are Dependent (i.e there is a relationship between them ) $$

How come the null and alternative tests are different in this case? Is it because there are no categorical variables? This is confusing knowing whether there will be a relationship or not depending on the type of variables, do you know what I mean?

davidgoes4wce · May 28, 2016

Re: University Statistics Discussion Marathon

$$ My little general rule that I tell students that I tutored$

$$ P-Value $ < $ Alpha $ \implies Accept H_{1}$

$$ P-Value $ > $ Alpha $ \implies Accept H_{0}$

$$ I try to keep things simple when deciding to Reject or Accept a Hypothesis$

BlueGas · May 28, 2016

Re: University Statistics Discussion Marathon

davidgoes4wce said:
$$ My little general rule that I tell students that I tutored at USyd, UNSW and UWS is if$

$$ P-Value $ < $ Alpha $ \implies Accept H_{1}$

$$ P-Value $ > $ Alpha $ \implies Accept H_{0}$

$$ I try to keep things simple when deciding to Reject or Accept a Hypothesis$

Yeah that's the easy part but what's the interpretation of the p value? That's the part which I need to understand. The two figures I posted (the one you helped me in and the second one now) have different null and alternative hypothesis tests.

davidgoes4wce · May 28, 2016

Re: University Statistics Discussion Marathon

BlueGas said:
How come the null and alternative tests are different in this case? Is it because there are no categorical variables? This is confusing knowing whether there will be a relationship or not depending on the type of variables, do you know what I mean?

I think if you read up a bit more on the 'Chi Squared of the Contingency table' it will be better for you. [/tex]

$$ In the case above you should have shown 4 different proportions. In that case you would have a Null Hypothesis: To test whether there are No differences among the four proportions nationally. In your alternate hypothesis, you want to test whether at least one of these four groups is significantly different.$

$A hypothesis you could test against could be along the lines of :$

$$ To test the null hypothesis that the proportions are equal:$
$H_{0}: \pi_{1}=\pi_{2}=\pi_{3}=\pi_{4}$

$$ Against the alternative that not all all four proportions are equal: (or if I reword it another way, at least one of these groups is not equally distributed)$
$H_{1}: $ Not all $ \pi_{j} $ are equal (where j =1,2,3,4) $$

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