The pulse rate (in bpm) of a random sample of 30 Peruvian Indians was collected. The mean pulse rate of the sample was 70.2 and the standard deviation was 10.51. Compute the 95% confidence interval for the population mean.

Respuesta :

lucic

Answer:

= 70.2 ± 3.761  bpm

Step-by-step explanation:

The question is on calculating the confidence interval for a population mean

The general expression is

CI = x ± z * δ/√n   where;

CI = confidence interval,

 x = mean of sample,

 δ = standard deviation,

  n= is sample size

   z = z* value from standard normal distribution according to   confidence level given.

Given that;

n= 30   x =70.2    δ=10.51          z* for 95% CI = 1.96

Then applying the expression

CI = x ± z * δ/√n

[tex]=\sqrt{n} = \sqrt{30} =5.477\\\\=\frac{10.51}{5.477} =1.919*1.96=3.761\\\\[/tex]

Cl = 70.2±3.761

= 70.2 ± 3.761  bpm