Answer: (84.876, 87.124)
Explanation:
Confidence interval for population mean if population standard deviation is unknown:
![\overline{x}\pm t_(\alpha/2)((s)/(√(n)))](https://img.qammunity.org/2021/formulas/mathematics/college/15i2rvzdtg6657cujaz48uhb2hrlp4cev0.png)
, where n= sample size
s= sample standard deviation
= sample mean
significance level
= critical-t value
Given: n= 64
Degree of freedom = n-1 = 63
s= 4.5
= 86
0.05
= 1.9983
Now, the required 95% confidence interval would be:
![86\pm (1.9983)((4.5)/(√(64)))\\\\=86\pm (1.9983)((4.5)/(8))\\\\=86\pm (1.9983)(0.5625)\\\\\approx86\pm 1.1240\\\\ =(86-1.1240,\ 86+1.1240)\\\\=(84.876,\ 87.124)](https://img.qammunity.org/2021/formulas/mathematics/college/3ufp8vl1lp6hymwqdvqqw4408bzqxdd9hg.png)
The 95% confidence interval for the population mean μ is between: (84.876, 87.124)