Respuesta :
Answer:
a) standard error of the mean =10.06
b) The margin of error = 17.3982
c) 95% of confidence intervals are
[tex](89.6018 ,124.3982)[/tex]
d) Lower limit of 95% of confidence interval = 89.6018
upper limit of 95% of confidence interval = 124.3982
The Population mean is lies between in these intervals
Step-by-step explanation:
Step(i):-
Given sample size 'n' = 20
Given sample mean was found to be 107 bpm with a standard deviation of 45 bpm.
Sample mean [tex]x^{-} = 107 bpm[/tex]
Sample standard deviation (S) = 45 bpm
a) standard error of the mean is determined by
[tex]S.E = \frac{S}{\sqrt{n} } = \frac{45}{\sqrt{20} }[/tex]
S.E = 10.06
b) The margin of error is determined by
[tex]M.E = \frac{t_{\alpha } X S }{\sqrt{n} }[/tex]
The degrees of freedom ν [tex]= n-1 =20-1=19[/tex]
[tex]t_{\alpha } = 1.729[/tex]
[tex]M.E = \frac{1.729X 45}{\sqrt{20} }[/tex]
[tex]M.E = \frac{77.805 }{4.472 } = 17.3982[/tex]
c) 95% of confidence intervals are determined by
[tex](x^{-} - t_{\alpha } \frac{S}{\sqrt{n} } , x^{-} + t_{\alpha } \frac{S}{\sqrt{n} } )[/tex]
[tex](107 - 1.729\frac{45}{\sqrt{20} } , 107 + 1.729\frac{45}{\sqrt{20} } )[/tex]
[tex](107 - 17.3982 } , 107 +17.3982 )[/tex]
[tex](89.6018 ,124.3982)[/tex]
d)
Lower limit of 95% of confidence interval = 89.6018
upper limit of 95% of confidence interval = 124.3982
The Population mean is lies between in these intervals
