Answer: (82.66, 87.34).
Explanation:
When population standard deviation is unknown and sample size is small , then the formula is used to find the confidence interval for
is given by :-
![\overline{x}\pm t^*(s)/(√(n))](https://img.qammunity.org/2021/formulas/mathematics/high-school/4zikiokgw1dwp0veedleuggjz9o7f8lsg4.png)
, where n = sample size ,
= sample mean , t*= two tailed critical value s= sample population standard deviation, .
Given,
, s=5, n=20 , degree of freedom = 19 [∵df=n-1]
For 95% confidence level ,
![\alpha=0.05](https://img.qammunity.org/2021/formulas/mathematics/middle-school/jq2szalj1kgt7x5aeuif4qgouxi8l1my6r.png)
By t-distribution table ,
t-value for
(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 :
![85\pm (2.0930)(5)/(√(20))](https://img.qammunity.org/2021/formulas/mathematics/college/cr41xh149r1ftarbe68r4gpapz43dv20z0.png)
![=85\pm (2.0930)(1.118034)](https://img.qammunity.org/2021/formulas/mathematics/college/3vdd8bgi1zfap1h8oime4mpfxynre2cgs9.png)
![\approx85\pm 2.34](https://img.qammunity.org/2021/formulas/mathematics/college/rns6wl8xakkq9stgz0f1tn8dcv5rb9fto5.png)
![=(85- 2.34,\ 85+2.34)\\\\=(82.66,\ 87.34)](https://img.qammunity.org/2021/formulas/mathematics/college/ctq1ustk3hd18qwyy4qq02tfbh2gdqg80r.png)
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).