Question 1201601
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<font color=red>Answers:</font>
(a) See explanation below.
(b) 0.00023 (approximate)
(c) n = 96 (approximate)


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Explanation:


Part(a)


Right-skewed means the main cluster of individuals are to the left while there are a few large outliers to the right (thereby forming a long stretched tail to the right).
In this context, it means that the vast majority of guinea pigs with cancer unfortunately do not survive too long after treatment. A few lucky individuals that are large outliers do survive for quite a while longer.
Another term for "right-skewed" is "positively skewed".


n = 49 = sample size
Because n > 30 is the case, the xbar distribution of sample means will be approximately normally distributed even if the population distribution isn't symmetric.
For more information, check out the Central Limit Theorem.


See this article for more detail
<a href="https://www.statology.org/central-limit-theorem/">https://www.statology.org/central-limit-theorem/</a>
Quote from the page: 
<pre>"If the population distribution is skewed, generally a sample size of at least 30 is needed.
If the population distribution is extremely skewed, then a sample size of 40 or higher may be necessary"</pre>

This article
<a href="https://www.whatissixsigma.net/confidence-intervals-why-n30-is-acceptable-as-population-representative/">https://www.whatissixsigma.net/confidence-intervals-why-n30-is-acceptable-as-population-representative/</a>
talks about why the number 30 is so special when it comes to the n > 30 criteria.


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Part(b)


mu = 100 = population mean survival time
sigma = 10 = population standard deviation of the survival times (tells us how spread out the population is)
n = 49 = sample size
xbar = sample mean survival time
Each survival time is measured in days.


We wish to compute 
P(xbar < 95)


Let's find the standard error
SE = Standard error
SE = sigma/sqrt(n)
SE = 10/sqrt(49)
SE = 1.428571 which is approximate


Now determine the z score when xbar = 95
z = (xbar - mu)/SE
z = (95 - 100)/1.428571
z = -3.50000105000031
z = -3.50
The task of finding P(xbar < 95) is equivalent to P(Z < -3.50)


I'm rounding to two decimal places so I can then use a table such as this one
<a href = "https://www.ztable.net/">https://www.ztable.net/</a>
A similar table can be found in the back of your stats textbook.


Use a table to find that:
P(Z < -3.50) = 0.00023
this leads back to
P(xbar < 95) = 0.00023


There is roughly a 0.023% chance of getting a sample mean survival time of less than 95 days.


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Part(c)


mu = 100 = population mean survival time
Add and subtract off 2 from the population mean
mu-2 = 100-2 = 98
mu+2 = 100+2 = 102


The goal is to find a sample size n that will lead to this statement
P(98 < xbar < 102) = 0.95


Let's find the standard error
SE = Standard error
SE = sigma/sqrt(n)
SE = 10/sqrt(n)
Right now we don't know what n is, so we leave the SE as shown above.


Now determine the z score when xbar = 98
z = (xbar - mu)/SE
z = (98 - 100)/(10/sqrt(n))
z = -2/(10/sqrt(n))
z = -2*sqrt(n)/10
z = -sqrt(n)/5
Repeat for xbar = 102 and you should get: z = sqrt(n)/5


P(98 < xbar < 102) 
is the same as 
P(-sqrt(n)/5 < Z < sqrt(n)/5)
which is in the format
P(-k < Z < k)


We need to find a value of k such that
P(-k < Z < k) = 0.95


You could use the table in reverse but it would take a lot of trial and error.
A better approach is to use a calculator such as this one
<a href="https://davidmlane.com/normal.html">https://davidmlane.com/normal.html</a>
or you could use a TI84 (or similar).


You should find that k = 1.96 approximately
P(-1.96 < Z < 1.96) = 0.95 approximately


We can now determine n.
k = 1.96
sqrt(n)/5 = 1.96
sqrt(n) = 5*1.96
sqrt(n) = 9.8
n = (9.8)^2
n = 96.04
n = 96


If n = 96, then the SE is
SE = Standard error
SE = sigma/sqrt(n)
SE = 10/sqrt(96)
SE = 1.020621 which is approximate


Now determine the z score when xbar = 98
z = (xbar - mu)/SE
z = (98 - 100)/1.020621
z = -1.95959126845322
z = -1.96


Repeat for xbar = 102
z = (xbar - mu)/SE
z = (102 - 100)/1.020621
z = 1.95959126845322
z = 1.96


Then you should find that P(-1.96 < z < 1.96) = 0.95 approximately. 
Use the table or a calculator. 
This helps confirm we have the correct value of n.


Conclusion: 
Having a sample size of n = 96 will lead to P(98 < xbar < 102) = 0.95 approximately.
Sampling 96 guinea pigs will have their sample mean survival time between 98 and 102 days with 95% probability.
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