Basic in Hypothesis Testing A-Level
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Dear alls
I'm the beginner in Hypothesis Testing. My teacher just gave me a note and let me study mysefl. I found Hypothesis Testing hard.
I guess I understand the most basic thing but when i red the example, it is so confusing.
Ex: Consider a right-tailed t-test with n =10 at 1% level of significance. From graphical calculator, InvT (0.99 , 9)= 2.821
Critical region :t>2.821
p- value = P (T>t) where t is the value of the test-static.
And my question is:
_ What is the meaning of 0.99 and 9. Even if i knew how to calculate that number, i don't have graphical calculator so how can I find 2.821
_ After i have critical region, how to find p ( which table and how to use it)
_ What is the meaning of T> t.
Thank you a lot. Plz help me.
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9 is the number of degrees of freedom. In the case of a single sample t-test, this is given by n-1 where n is the number of observations.
0.99 reflects the level of significance. It means that the chances of those observations having the values that you're testing has a 99% chance of arising from your hypothesis. In other words, there is a 1% chance that your observations came about completely by chance. 1%, or rather, 0.01 is the p-value. Mathematically, it is the probability that your t-value falls in the upper 1% of the hypothesis distribution. This is equivalent to the probability that your t-value T is bigger than the critical t-value t, hence T>t.
To find the critical t-value t, you can use this table: http://www.math.unb.ca/~knight/utility/t-table.htm for now, but you should ask your teacher for one. Look at the column (One-tail, 0.01) and row 9.
To calculate T, the t-value from your observations, first calculate the mean <m> and standard deviation s of your n observations. You will also need your hypothesis mean M.
Then T= ( <m>-M ) / ( s/root(n) )
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First of all, thank you for answering me. I get the first question. However the second and the third, I'm still confusing about it. The table which you gave me allows me to find 2.821. But about the p, how can i find it. Can you give me the detail fomula and number?? Plz make it as detail as possible.
Thank you one more
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Hi,
Are you taking H2 Maths? I am discussing these assuming that you are encountering these at A-level.
Ex: Consider a right-tailed t-test with n =10 at 1% level of significance.
From graphical calculator (GC), InvT (0.99 , 9)= 2.821.I'd suggest that you follow these steps to visualise:
1. Draw a bell-shaped curve.2. Shade the right-end of the curve. The shaded region has an area of 0.01 or 1%.
3. On the x-axis of this curve, the t-value that gives the area is 2.821.
So we can say that:
1. the critical region refers to the set of t-values such that t > 2.821.
2. the probability that T > 2.821 is equal to 0.01, or that P(T > 2.821) = 0.01, we call 0.01 the p-value.
Given any t-value, we can use TI-GC to find p-value. Take, for example, the p-value corresponding to P(T > 1.89) at 9 degrees of freedom:
1. Press 2ND-VARS.
2. Select 6: tcdf(
3. Enter 1.89 for "lower:", enter E99 for "upper:", enter 9 for "df:".
4. Obtain the p-value of 0.0457, to 3 s.f.
You can use the table in MF15 to find the t-value for specific levels of significance. Using your original example:
1. Locate the column with p = 0.99.
2. Locate the row with degrees of freedom = 9.
3. Obtain the t-value of 2.821.
Thanks.
Cheers,
Wen Shih -
Thank ...I'm now taking H2 maths. The only calculator that i have is CASIO fx- 570ES PLUS. Can you guide me by that computer or table?
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I guess a example with concrete number will help me better.
Experience has shown that the scores obtained in a particular test are normallt distributed with mean score 70 and variance 36. When the test is taken by a ramdom sample of 36 students, the mean score is 68.5. Is it sufficient evidence, at 3% level of significance, that these student have not performend as well as expected?
Solution from my book:
_ Let X be the random variable for the score of particular test and x (I change the signal so that I can write in here) be its population mean.
Ho = x =70
H1 = x<70
Under Ho , X avg - N ( x, variance) => X avg -N( 70, 36/36)
Since the sample is from a nornal population with known variance, we can use z- test. ( Is it z-test table http://www.stattools.net/zTest_Tab.php).
Test statistic , Z= ( X avg- 70)/sqr of 1. ( I understant this step)
equal N (0.1 ) ( Why like this ???)
At 3% significance, we perform a 1 tail test. From graphic calcularor, critical region z <-1.881 ( How can do this without graphic calculator).
We reject Ho if z<-1.881
The value of test statistic z =-1.5
Since Z> -1.881 we do not reject Ho.
Can you help me those question which I wrote in red.Thank you
P/s : I don't see any function of z -test here.
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Hi,
Let X be the random variable representing test scores.
Let mu be the population mean score.
H_0 : mu = 70
H_1 : mu < 70
Level of significance: 3%
Under H_0, X-bar ~ N(70, 1).
We use the Z-test, with the help of GC if you wish to use it.
Test statistic Z = (X-bar - 70) / sqrt(1) ~ N(0, 1), because of standardisation.
To obtain the critical region, use of GC is necessary, because the Normal distribution table in MF15 does not have p = 0.97. The nearest value of p is 0.975.
To use Casio-GC, the steps are:
1. Go to STAT.
2. Select DIST.
3. Select NORM.
4. Select InvN.
5. Select Left for Tail.
6. Enter 0.03 for Area.
7. Enter 1 for sigma.
8. Enter 0 for mu.
You will obtain -1.881, correct to 3 decimal places.
Suppose the level of significance is 5%, then we can conveniently use the Normal table by considering p = 0.95 and then the required z-value will be -1.645, by symmetry of the Normal distribution.
Hope these help.
Cheers,
Wen Shih -
My generation never touched a graphic calculator even once.
And I'm not old at all.
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Hi,
In your example, the calculated z-value = -1.5. What is the p-value?
If we want to find the p-value, which is P(Z < -1.50), we can use the Normal table as follows:
1. Go to the row with z = 1.5.
2. For the same row, move to the next column to obtain the value 0.9332.
Now P(Z < 1.50) = 0.9332, then
P(Z < -1.50) = P(Z > 1.50)
= 1 - 0.9332
= 0.0668.
Thanks.
Cheers,
Wen Shih -
Originally posted by SBS2601D:
My generation never touched a graphic calculator even once.
And I'm not old at all.
Hi,
There was no GC when I was a student too. Much was learnt from first principles and use of statistics tables.
Cheers,
Wen Shih -
Guys, i guess in my example, the key feature is finding critical area. My problem is how to find it. I don't have Graphic calculator. Someone, plz, tell me how to solve this by table. And plz give me the table too
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And base on the solution of my book, i think there is no need to count p value. Just compare Z to critical value
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You must understand the normal curve and understand that the area under the curve to the left of the hypothesised standardised mean (between 0 and 1) represents the probability that the value of whatever you measure is less than the hypothesised mean.
The table is widely available on internet....just google Z table.
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Hmmmm
The table that you said is the table which we use in normal distribution, isn't it??? But how can I use it in hypothesis testing.
At 3% significance, we perform a 1 tail test. From graphic calcularor, critical region z <-1.881
Is it the step which we need to use table.
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Those values are the said z values.
You must know also that the z value is the standardised forms of the hypothesised means.
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Hope this video lecture on hypothesis testing will be of help.
http://www.learnerstv.com/video/Free-video-Lecture-11986-Maths.htm#
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Originally posted by Theplayfulgirl brainy:
Hmmmm
The table that you said is the table which we use in normal distribution, isn't it??? But how can I use it in hypothesis testing.
At 3% significance, we perform a 1 tail test. From graphic calcularor, critical region z <-1.881
Is it the step which we need to use table.
Hi,
Yes, the table is in MF15. Please refer to page 6 of the document found at:
http://www.seab.gov.sg/aLevel/2013Syllabus/ListMF15.pdf
If the level of significance is 3%, you need to use GC to find the z-value. Do note that GC is a must for H2 Maths, especially when it comes to solving of probability and statistics questions.
The table is only convenient when it comes to certain values of level of significance, e.g., 10%, 5%, 2.5%, 1%, etc. The table is also useful when we want to find the p-values corresponding to certain z-values.
I'd encourage you to clarify lingering doubts with your teacher. Watching videos can be helpful but clarifying in person when he/she is in close proximity will be most beneficial, in my view.
Thanks, jiayou!
Cheers,
Wen Shih -
Thanks you alls, i guess i get it now :)
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How to use TI 83 to find z value?? I bought a GC now but i don't know how to use it :)
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Originally posted by Theplayfulgirl brainy:
How to use TI 83 to find z value?? I bought a GC now but i don't know how to use it :)
Hi,
1. Press 2ND, VARS.
2. Select invNorm.
3. Enter 0.95, 0, 1. This means we want to find the value of z for which P(Z <= z) = 0.95.
4. Obtain 1.6449 as the output from GC.
Thanks.
Cheers,
Wen Shih -
... Hurahh. I did it.
Anw, I met a problem with this exercise of Hypo:
For a random sample of 10 observations of normal variable X, the mean is x' where x'=4.344
sigma (x-x')^2 =0.8022
Using a 5% significance level, carry out a 2 tailed test whether the mean of X is 4.58
Ho: mu=mu o= 4.58
H1 :mu doesn't equal mu o
I did the same way with my book except one step.
My solution : Because of 5% significance level, reject Ho if z< -1.9599 or z>1,9599
Book solution: 5% critical value, t 0,025 ; 9 =2.262
..................
If |t|> t 0.025; 9 reject Ho
My question is
_Whether my solution is wrong and why?
_ What is t 0.025 ; 9.?
What is the meaning of 9?
How to find that value by TI 83?
Thank you
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Hi,
You used the Z-test which is inappropriate in this case. Since the data comes from a Normal population, n (sample size) is small and population variance is unknown, we will use the T-test instead.
t_{0.025} refers to the t-value that would give an area of 0.025. We consider 0.025 since we are looking at a 2-tailed test at 5% (0.05) level of significance.
9 refers to the degrees of freedom and this is calculated by the formula:
sample size - 1.
The required GC steps are
1. Press 2ND, VARS.
2. Select invT(
3. Enter 0.025, 9.
4. Obtain -2.262... as output.
Thanks.
Cheers,
Wen Shih -
For a random sample of 10 observations of normal variable X
So the word normal variable X decide if I'm able to use N test or not..
OMG. Those stuffs are so confusing.
@@
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Hi,
To use the T-test, we need to have the following conditions met:
1. Population is Normal or assumed to be Normal.
2. Sample size is small.
3. Population variance or standard deviation is unknown.
Thanks.
Cheers,
Wen Shih -
Thank you. I got it now