Answer:
90% Confidence Interval for the actual average weight of purses.
(4.7412 , 5.2588)
Step-by-step explanation:
Explanation:-
Given sample size 'n' = 60
mean of the sample x⁻ = 5 pounds
Standard deviation of the sample 'S' = 1.2 pounds
Level of significance = 0.10
90% Confidence Interval for the actual average weight of purses.
[tex](x^{-} - t_{\frac{\alpha }{2} } \frac{S}{\sqrt{n} } , x^{-} +t_{\frac{\alpha }{2} } \frac{S}{\sqrt{n} } )[/tex]
[tex](5 - 1.6711\frac{1.2}{\sqrt{60} } , 5 +1.671\frac{1.2}{\sqrt{60} } )[/tex]
( 5 - 0.2588 , 5 + 0.2588)
90% Confidence Interval for the actual average weight of purses.
(4.7412 , 5.2588)
