We have to perform an hypothesis test for the mean.
The claim is that the patient's mean HC is greater than the population average.
Then, the null and alternative hypothesis are:
[tex]\begin{gathered} H_0:\mu=14.4 \\ H_a:\mu>14.4 \end{gathered}[/tex]The significance level is 0.05.
The sample has a size n = 10.
The sample mean is M = 15.6.
The standard deviation of the population is known and has a value of σ = 2.5.
We can calculate the standard error as:
[tex]\sigma_M=\frac{\sigma}{\sqrt{n}}=\frac{2.5}{\sqrt{10}}\approx0.7906[/tex]Then, we can calculate the z-statistic as:
[tex]z=\frac{M-\mu}{\sigma_M}=\frac{15.6-14.4}{0.7906}=\frac{1.2}{0.7906}\approx1.52[/tex]This test is a right-tailed test, so the P-value for this test is calculated as:
[tex]P-value=P(z>1.52)=0.064[/tex]As the P-value (0.06) is bigger than the significance level (0.05), the effect is not significant.
The null hypothesis failed to be rejected.
At a significance level of 0.05, there is not enough evidence to support the claim that the patient's mean HC is greater than the population average.
Answer:
z = 1.52
P-value = 0.064
Conclusion: fail to reject the null hypothesis.
