Answer: (82.66, 87.34).
Step-by-step explanation:
When population standard deviation is unknown and sample size is small , then the formula is used to find the confidence interval for [tex]\mu[/tex] is given by :-
[tex]\overline{x}\pm t^*\dfrac{s}{\sqrt{n}}[/tex]
, where n = sample size , [tex]\overline{x}[/tex]= sample mean , t*= two tailed critical value s= sample population standard deviation, .
Given, [tex]\overline{x}=85[/tex], s=5, n=20 , degree of freedom = 19 [∵df=n-1]
For 95% confidence level , [tex]\alpha=0.05[/tex]
By t-distribution table ,
t-value for [tex]\alpha/2=0.025[/tex] (two tailed) and df =19 is t*=2.0930
Now , the 95% confidence interval for the mean heart rate of adults in the population will be :
[tex]85\pm (2.0930)\dfrac{5}{\sqrt{20}}[/tex]
[tex]=85\pm (2.0930)(1.118034)[/tex]
[tex]\approx85\pm 2.34[/tex]
[tex]=(85- 2.34,\ 85+2.34)\\\\=(82.66,\ 87.34)[/tex]
Hence, the required interval is (82.66, 87.34).
Interpretation : A person can be 95% confident that the mean heart rate of adults in the population lies between (82.66, 87.34).
