Question 1207829
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Please ask only one question per post.
I'll do problem 1 to get you started.


n = 20 = sample size
mu = 37 = the claimed population mean body temperature
xbar = 36.78 = sample mean body temperature
s = 0.35 = sample standard deviation of the temperatures
sigma = population standard deviation = unknown
alpha = level of significance = 0.05


Null Hypothesis: mu = 37
Alternative Hypothesis: mu =/= 37
The claim is made in the null. 
The alternative hypothesis indicates we have a two-tailed test.


We don't know the value of sigma and n > 30 is not the case, so we must use the T distribution.
df = degrees of freedom
df = n-1
df = 20-1
df = 19


Use a T table such as this one
<a href="https://www.sjsu.edu/faculty/gerstman/StatPrimer/t-table.pdf">https://www.sjsu.edu/faculty/gerstman/StatPrimer/t-table.pdf</a>
Such a table can be found at the back of your stats textbook.
Highlight the df = 19 row and the column that mentions "two tails = 0.05" since this is the alpha level.
The intersection of this row and column yields the approximate t critical value t = 2.093


What does this tell us?
It tells us that P(-2.093 < t < 2.093) = 0.95 approximately when df = 19.
The 0.95 is the area of the main body while 1-0.95 = 0.05 is the combined area of the two tails. 


If the test statistic is in the interval -2.093 < t < 2.093, then we do not reject the null.
Otherwise, we'll reject the null.


Let's compute the test statistic.
testStatistic = (xbar - mu)/( s/sqrt(n) )
testStatistic = (36.78 - 37)/( 0.35/sqrt(20) )
testStatistic = -2.811 approximately


The value t = -2.811 is NOT in the interval -2.093 < t < 2.093
This value is in the rejection region. It's to the left of the interval mentioned.
Therefore we'll reject the null and conclude that mu =/= 37 must be the case.
The claim "the mean body temperature of normal and healthy adults is equal to 37°C" appears to be false.
Either mu > 37 or mu < 37.


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Another approach using p-value


We calculated the test statistic to be roughly -2.811
Let's find the value of P(t < -2.811)
Use a stats calculator such as a TI83 to determine that P(t < -2.811) = 0.00558 approximately when df = 19.


We're doing a two-tailed test, so don't forget to double that result and you would get 2*0.00558 = 0.01116
This is the approximate p-value.
It is smaller than alpha (0.05), so we must reject the null. 


A useful phrase to remember: "If the p-value is low, then the null must go" which would translate to "if the p-value is smaller than alpha, reject the null".
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