# Thread: Statistics Marathon & Questions

1. ## Re: University Statistics Discussion Marathon

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

2. ## 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.

3. ## Re: University Statistics Discussion Marathon

Am I correct for this?

4. ## Re: University Statistics Discussion Marathon

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

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

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

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

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

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

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

$The test statistic is 210.9$

$Number of rows r=5$

$Number of columns c=4$

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

$Using excel function the P-Value is ,$

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

$=0.0000$

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

5. ## Re: University Statistics Discussion Marathon

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

6. ## Re: University Statistics Discussion Marathon

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

7. ## Re: University Statistics Discussion Marathon

Originally Posted by davidgoes4wce
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?

8. ## 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".

9. ## Re: University Statistics Discussion Marathon

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

10. ## Re: University Statistics Discussion Marathon

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

11. ## Re: University Statistics Discussion Marathon

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

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

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

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

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

12. ## Re: University Statistics Discussion Marathon

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

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

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

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

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

13. ## Re: University Statistics Discussion Marathon

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

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

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

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

14. ## Re: University Statistics Discussion Marathon

Originally Posted by davidgoes4wce
$My little general rule that I tell students that I tutored at USyd, UNSW and UWS is if$

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

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

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

15. ## Re: University Statistics Discussion Marathon

Originally Posted by BlueGas
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]

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

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

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

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

16. ## Re: University Statistics Discussion Marathon

Speaking of Statistics have any of you guys used R Programming and Statistical software? (for say work or university purposes)?

I want to be more proficient at this software and I enjoy using it in comparison to Excel, SPSS or Minitab for that matter as the speed is alot quicker.

17. ## University Statistics Discussion Marathon

Originally Posted by BlueGas
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.
The p-value is the probability of an observation occurring assuming the null hypothesis is actually true. If the p-value is very low then the observation you see is very unlikely if the null hypothesis is true. If you believe the observation you see is typical (has a higher probability of occurring) then it is likely that your assumption may not be right, hence the decision reject it.

Originally Posted by davidgoes4wce
Speaking of Statistics have any of you guys used R Programming and Statistical software? (for say work or university purposes)?

I want to be more proficient at this software and I enjoy using it in comparison to Excel, SPSS or Minitab for that matter as the speed is alot quicker.
I've used R once upon a time

18. ## Re: University Statistics Discussion Marathon

Originally Posted by Trebla
The p-value is the probability of an observation occurring assuming the null hypothesis is actually true. If the p-value is very low then the observation you see is very unlikely if the null hypothesis is true. If you believe the observation you see is typical (has a higher probability of occurring) then it is likely that your assumption may not be right, hence the decision reject it.

I've used R once upon a time
There seems to be alot of jobs out there for people who are skilled up in R and SPSS. It pays to focus on one or two specific areas.

19. ## Re: University Statistics Discussion Marathon

Topic of this question is Sum and Difference of independent normal variates

Is there a mistake in this question? I get the feeling that they have substituted the wrong value of 40 in , which should have been 20.
$In this question there is an assumption that \mu=20 and a \sigma=4$

20. ## Re: University Statistics Discussion Marathon

$According to a recent poll, 29 \% of adults in a certain area have high levels of cholesterol. They report that such elevated levels could be financially devasting to the regions healthcare system and are a major concern to health insurance providers. According to recent studies, cholesterol levels in healthy adults from the area average about 210 mg/DL, with a standard deviation of 20 mg/DL, and are roughly Normally distributed. If the cholesterol levels of a sample of 47 healthy adults from the region is taken, answers part (a) through (d).$

$(a) MCQ. What shape will the sampling distribution of the mean have?$

$a) The sampling distribution of the mean is skewed right$
$b) The sampling distribution of the mean is skewed left$
$c) The sampling distribution of the mean is normal$
$d) There is not enough information to determine the shape of the sampling distribution$

$(b) What is the mean of the sampling distribution ?$

$(c) What is the standard deviation?$

$(d) If the sampling distribution were increased to 90, how would your answers to parts (a) - (c) change?$

$for part (a), the shape of the distribution would? a = also be normally distributed, b = become skewed left, c = become skewed right$

$- for part (b), the mean of the sampling distribution would? a = change to, b = remain as .... if it is suppose to change, what is the new sampling distribution?$

$-for part (c), the standard deviation of the sampling distribution would? a = change to, b = remain as .... if it is suppose to change, what is the new standard deviation?$

21. ## Re: University Statistics Discussion Marathon

Originally Posted by davidgoes4wce
$According to a recent poll, 29 \% of adults in a certain area have high levels of cholesterol. They report that such elevated levels could be financially devasting to the regions healthcare system and are a major concern to health insurance providers. According to recent studies, cholesterol levels in healthy adults from the area average about 210 mg/DL, with a standard deviation of 20 mg/DL, and are roughly Normally distributed. If the cholesterol levels of a sample of 47 healthy adults from the region is taken, answers part (a) through (d).$

$(a) MCQ. What shape will the sampling distribution of the mean have?$

$a) The sampling distribution of the mean is skewed right$
$b) The sampling distribution of the mean is skewed left$
$c) The sampling distribution of the mean is normal$
$d) There is not enough information to determine the shape of the sampling distribution$

$(b) What is the mean of the sampling distribution ?$

$(c) What is the standard deviation?$

$(d) If the sampling distribution were increased to 90, how would your answers to parts (a) - (c) change?$

$for part (a), the shape of the distribution would? a = also be normally distributed, b = become skewed left, c = become skewed right$

$- for part (b), the mean of the sampling distribution would? a = change to, b = remain as .... if it is suppose to change, what is the new sampling distribution?$

$-for part (c), the standard deviation of the sampling distribution would? a = change to, b = remain as .... if it is suppose to change, what is the new standard deviation?$
$Here is my attempt at this question$

$(a) Shape will be bell shaped i.e. mean is normal$

$(b) Mean of sampling distribution is 210$

$(c) Standard deviation = \frac{\sigma}{\sqrt{n}}=\frac{20}{\sqrt{47}}=2.917$

$(d) If we increase the sample size to 90. Then the shape of distribution would be Normally Distributed.$

$Mean will Remain the same$

$Standard deviation = \frac{20}{\sqrt{90}}=2.108$

$Hence it will increase sample size then standard deviation will be decreased.$

22. ## Re: University Statistics Discussion Marathon

$An analyst collected data from 25 randomly selected transactions and found the purchase amounts​ (in \). The mean is \​29.92 and the standard deviation is \​14.40. The analyst wants to know if the mean purchase amount of all transactions is at least ​\24. What is the​ P-value of the test​ statistic ?$

$How would you type this code in Excel to determine the P-Value ?$

23. ## Re: University Statistics Discussion Marathon

I'll admit I got this question from a quiz and from my understanding reading a textbook (which shows Excel cell calculation) my interpretation of this calculation is it should be :

= T.DIST.RT((25.23-19)/14.03, 24)

$It's based on the t-test statistic formula t=\frac{\bar{x}-\mu}{{s} \div \sqrt{n}}$

$Degrees of Freedom in this case = 24 \ df$

24. ## Re: University Statistics Discussion Marathon

$It is recommended that adults get 8 hours of sleep each night. A researcher hypothesized college students get less than the recommended number of hours of sleep each night, on average. The researcher randomly sampled 20 college students and obtained a p-value of 0.10. Suppose the researcher sampled more college students and that the sample mean and sample standard deviation stayed the same. Would the P-Value be lower, be higher, or stay the same ?$

25. ## Re: University Statistics Discussion Marathon

Originally Posted by davidgoes4wce
$It is recommended that adults get 8 hours of sleep each night. A researcher hypothesized college students get less than the recommended number of hours of sleep each night, on average. The researcher randomly sampled 20 college students and obtained a p-value of 0.10. Suppose the researcher sampled more college students and that the sample mean and sample standard deviation stayed the same. Would the P-Value be lower, be higher, or stay the same ?$
I'm guessing that the p value would change but I don't know lol, would like to know the answer though.

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