Question 1171312
<font color=black size=3>
Hypothesis:
H0: mu = 4.8
H1: mu < 4.8
Left tailed test


alpha = 0.05


n = 6 is the sample size
Because n < 30 and sigma is not known, this means we must use a T test. Sigma is the population standard deviation.


While the population standard deviation is not known, the population mean (mu) is known. In this case, it is mu = 4.8, which is how the hypothesis is formed above.


For the female patient, the samples lead to
xbar = sample mean = 4.4
s = sample standard deviation = 0.28


The standard error (SE) is
SE = s/sqrt(n)
SE = 0.28/sqrt(6)
SE = 0.11430952132989


The T test statistic is
t = (xbar - mu)/(SE)
t = (4.4 - 4.8)/(0.11430952132989)
t = -3.49927106111857
t = -3.499271


Use a T cdf function on your calculator to find that 
P(T < -3.499271) = 0.0086 approximately when df = n-1 = 5
df = degrees of freedom
So the p value is approximately 0.0086
This p value is smaller than alpha = 0.05, so we reject the null and conclude that mu < 4.8


Another way to reject the null is to notice that the test statistic (-3.499271) is to the left of the T critical value (-2.015) when alpha = 0.05 and df = 5. Use a calculator or a T table to find the T critical value. So this test statistic is in the rejection region.


Interpretation: Her mean red blood cell count is less than 4.8, which means it's lower than normal.
</font>