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Sagot :
Answer:
95% confidence interval for the proportion of blocks that are sufficiently strong
(0.83913 , 1.02293)
Step-by-step explanation:
Step:-1
Given that the sample size 'n' = 29 blocks
Estimate proportion
[tex]p = \frac{x}{n} = \frac{27}{29} = 0.93103[/tex]
Step:-2
95% confidence interval for the proportion of blocks that are sufficiently strong
[tex](p^{-} -Z_{0.05} \sqrt{\frac{p(1-p)}{n} } , p^{-} + Z_{0.05} \sqrt{\frac{p(1-p)}{n} } )[/tex]
[tex](0.93103 -1.96 \sqrt{\frac{0.93103(1-0.93103)}{29} } , 0.93103 + 1.96 \sqrt{\frac{0.93103(1-0.93103)}{29} } )[/tex]
(0.93103 - 0.0919 , 0.93103 +0.0919)
(0.83913 , 1.02293)
Final answer:-
95% confidence interval for the proportion of blocks that are sufficiently strong
(0.83913 , 1.02293)
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