# Statistical test

#### Shailee

##### New Member
Null Hypothesis : Mean of Intra-subject Time to Haemostasis (TTH) = > 10 min
Alternate Hypothesis : Mean of Intra-subject Time to Haemostasis (TTH) </= 10 min

Sample Size : = 70

I have TTH values for all the 70 samples as performed during the study.

I calculated values for (TTH - 10) for all samples.

After this, I performed paired sample t test, at significance 0.05 , two tailed for values of (TTH) & (TTH-10) (https://www.socscistatistics.com/tests/ttestdependent/default.aspx)
The value of p is .01081. The result is significant at p < .05.
Please help to know if this means that the null hypothesis is rejected in favour of alternate hypothesis.

Last edited:

#### GretaGarbo

##### Human
I performed paired sample t test,
Where do the pairs comes from? No, I don't think this is correct.

Null Hypothesis : Mean of Intra-subject Time to Haemostasis (TTH) = > 10 min
You can just do:

t = (mean(TTH) - 10)/(s/sqrt(70))

where s is the standard deviation.

If the t is less than -1.64 then the null hypothesis will be rejected. Example: t= -2.5 will reject the null hypothesis.

That is for a one sided test. For a two sided test the absolute value of t needs to be larger than 2.

#### Shailee

##### New Member
I am sharing the calculation.

Treatment 1 = TTH for all 70 samples
Treatment 2 = TTH - 10 for all 70 samples.

#### Attachments

• 100.8 KB Views: 3

#### Shailee

##### New Member
I have used attached reference for my study.
If you can, please help to know how analysis was performed, I can perform my analysis accordingly.

#### Attachments

• 99.9 KB Views: 1
• 64.5 KB Views: 1

#### GretaGarbo

##### Human
I don't understand this. You used a pairwise test. Did the patients go twice with the ambulance?

The numbers in the first slide does not match the numbers in the next. Also there it sayes n= 167 not 70.

The much lower median as compared to the mean indicates a very skewed distribution or outliers.

#### Shailee

##### New Member
Apologies for the confusion.

I am sharing a reference study, I am trying to run my study in a similar way.

Can you help to know what data was collected and how analysis is performed for the attached reference study.

#### Attachments

• 99.9 KB Views: 1
• 64.5 KB Views: 1

#### GretaGarbo

##### Human
I am sorry but I am not so good in English.

What is "Haemostasis "? Does it have something to do with blod?

Waht is "ambulation"? Does it have something to do with walking? (I thought it was about the ambulance!

What is the sample size? Is it 70? Or much larger?

Where does the data comes from? Have you done the measurements and created them yourself?

What is your reasearch question? What is it you want to investigate? What is it you want to conclude?

Try to answer these with very simple English.

#### Shailee

##### New Member
Research Question: How much time does it take to stop bleeding (Haemostasis) after transracial catheterisation.

Study product: Chitosan dressing used to stop bleeding from puncture site

Method : Chitosan dressing was applied on the puncture site to stop bleeding.

Sample Size: 70

I have data for each patient:
Patient 1 : time to stop bleeding = 4 min
Patient 2: time to stop bleeding = 7 min
Patient 3: ...
....
...
Patient 70: time to stop bleeding = 5 min

Mean time to stop bleeding = 5.43 +/- 1.36 minutes.

I want to conclude that this mean time to stop bleeding is lesser as compared to standard ( as per literature standard time tostados top bleeding is 10 minutes)

Let me know if further information is required.

#### GretaGarbo

##### Human
Now it is more clear. There is no pairing.

You just have the mean and need to do a confidence interval.

So if "1.36" is the standard deviation( s), then you will get that a confidence interval this way:

mean +/- t*s/sqrt(n)

where "t" comes from the t-distribution and gives a 95% confidence interval. t=1.99 here. sqrt(n) means the square root of n and n is 70.

A 95% confidence interval:

5.43 +/- 1.99*1.36/sqrt(70)

gives:
[5.106523 ; 5.753477]

Since the confidence interval does not include "10" the hypothesis that the population mean is equal to 10 is rejected.

(Dou you prefer a t-test? It would reject the null hypotesis with a very low p-value (1.633491e-39))

There might be a skewness in your data since the mean is larger than the median. But with 70 observation the confidence interval will be approximately valid by the central limit theorem.

#### Shailee

##### New Member
Thanks for your valuable inputs.

How do I perform a t test and get a p value to reject the null hypothesis.

I was referring to that attached reference.
Attached reference has ambulation time. and has a p value of <0.0001 to reject the null hypothesis.

In my case I am trying to check similar analysis with time to stop bleeding.

If you can, do let me know how data might be collected and testes performed for attached reference and if I can use similar theories for my study for time to stop bleeding.

#### Attachments

• 99.9 KB Views: 3
• 64.5 KB Views: 3

#### GretaGarbo

##### Human
mean +/- t*s/sqrt(n)
A 95% confidence interval:

5.43 +/- 1.99*1.36/sqrt(70)

gives:
[5.106523 ; 5.753477]
A t-test and a confidence interval is very similar. Actually there is a correspondence between t-test and confidence intervals (That is called the correspondence theorem and is written in Casella-Berger for the more statistical oriented people.)

A t-test. You can do it like this:

t_obs = (mean - 10)/(s/sqrt(n))

Now, if the observed t-value (t_obs) is larger in absolute value that the critical t-value of 1.99, then it is statistically significant.

Whith your values it would be:

t_obs = (5.43 - 10) /(1.36/sqrt(70)) = -28.11424

And that is really larger in absolute value than 1.99. Its p-value would be 1.633491e-39, which is very close to zero.

This is based on my assumption that the standarddeviation is 1.36.

I just did this little thing in R (you can download R for free)

Code:
mn <- 5.43
std <- 1.36
n   <- 70
qt(0.975, n-1)
mn + c(-1 , +1)*1.99*std/sqrt(n)

(mn -10)/(std/sqrt(n))
2*pt(-28.11, 70-1)

#### GretaGarbo

##### Human
About the attached reference:

Let's see now: Inter-national is between nations. So I believe that inter-personal is between persons. But an intra-net is within a company. So I assume that an intra-individual will be within the individual.

But why measure within the individual? Isn't the question if the mean between individuals will be ca: 10 minutes?

"The intra subject difference... was analysed with a paired t-test". I don't understand this. If you take the paired difference of an individual it will be about 0. Or is there an treatent of before-treatment and after-treatmen?

#### Shailee

##### New Member
Thanks for your inputs. If t test is similar to Confidence Interval, I will just go ahead with confidence interval.

Also is it possible to find confidence interval for below outcome.

Patient comfort levels as assessed by verbal inquiry:
Excellent - 59 patients
Good - 11 patients
Fair - 0 patients
Not acceptable 0 patients.

#### Shailee

##### New Member
About the attached reference:

Let's see now: Inter-national is between nations. So I believe that inter-personal is between persons. But an intra-net is within a company. So I assume that an intra-individual will be within the individual.

But why measure within the individual? Isn't the question if the mean between individuals will be ca: 10 minutes?

"The intra subject difference... was analysed with a paired t-test". I don't understand this. If you take the paired difference of an individual it will be about 0. Or is there an treatent of before-treatment and after-treatmen?
I don't understand how the test was performed, this was available in literature and so I was referring to the same.

#### GretaGarbo

##### Human
I don't understand how the test was performed, this was available in literature and so I was referring to the same.
I don't understand that either. You must know more about that than I do. Does it make any sense to do a pairwise test?

- - - -

Also is it possible to find confidence interval for below outcome.

Patient comfort levels as assessed by verbal inquiry:
Excellent - 59 patients
Good - 11 patients
Fair - 0 patients
Not acceptable 0 patients.
You could do a confidence interval for the proportion "excellent" and "good"

- - -

Also I note that you have written in several threads about the same problems. Please don't do that. Now, both I and @hlsmith have tried to answer the same question.

#### hlsmith

##### Less is more. Stay pure. Stay poor.
Just provide a link to the paper you keep referencing. That would clear up the issues. I am assuming the study design for the referenced study is different from what you have done or they just did the wrong analyses or reported it incorrectly (which isn't uncommon).