Respuesta :
Answer:
Yes, it is appropriate to use Student's t method because we don't have any information about the population standard deviation.
Step-by-step explanation:
We are given that in an experiment to study the regulation of insulin secretion, blood samples were obtained from seven dogs before and after electrical stimulation of the vagus nerve.
The following values show the increase in the immunoreactive insulin concentration (µU/ml) in pancreatic venous plasma;
30, 100, 60, 30, 130, 1,060, 30.
Also, it is given that Student's t method yields the following 95% confidence interval for the population mean: -145< m <556.
Firstly, here it is appropriate to use Student's t method because we don't have any information about the population standard deviation.
So, the pivotal quantity for a 95% confidence interval for the population mean is given as;
P.Q. = [tex]\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }[/tex] ~ [tex]t_n_-_1[/tex]
where, [tex]\bar X[/tex] = sample mean increase in the immunoreactive insulin concentration = [tex]\frac{30 +100 +60+ 30+ 130+ 1,060+ 30}{7}[/tex] = 205.71
s = sample standard deviation = [tex]\sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} }[/tex] = 378.72
n = sample of dogs = 7
Also, [tex]t_(_\frac{\alpha}{2}_)[/tex] at [tex]\alpha[/tex] = 5% with (7-1 = 6) degree of freedom is 2.447 from t-table.
So, 95% confidence interval for the population mean ([tex]\mu[/tex]) is given by :
= [tex]\bar X \pm t_(_\frac{\alpha}{2}_) \times \frac{s}{\sqrt{n} }[/tex]
= [tex]205.71\pm 2.447\times \frac{378.72}{\sqrt{7} }[/tex]
= [205.71 - 350.27 , 205.71 + 350.27]
= [-145 , 556]
Hence, 95% confidence interval for the population mean is [-145, 556].
