Can someone please help me
1 answer:
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Answer:
The sample size should you gather to achieve a 0.45 hour margin of error
(n) = 411
Step-by-step explanation:
<u>Step(i)</u>
Given data a preliminary sample of 35 bacteria reveals a sample mean of x = 74 with a standard deviation of s = 5.4
Given the margin of error = 0.45
The degrees of freedom = n-1 = 35-1=34
The 90% of level of significance of t- distribution
t₀.₁₀ = 1.69 ( from table at 34 degrees of freedom at 0.10 level of significance)
<u>Step(ii)</u>
Margin of error = 1.69S / √n
Given data sample standard deviation S =5.4 hours.
margin of error = o.45
Margin of error =
use this formula to determine the sample size
√n = 1.69X5.4/0.45
√n = 20.28
squaring on both sides n= 411.27≅411
<u>Conclusion</u>:-
The sample size should you gather to achieve a 0.45 hour margin of error
(n) = 411
