Question

You are a researcher studying the lifespan of a certain species of bacteria. A preliminary sample of 35 bacteria reveals a sample mean of ¯ x = 74 x¯=74 hours with a standard deviation of s = 5.4 s=5.4 hours. You would like to estimate the mean lifespan for this species of bacteria to within a margin of error of 0.45 hours at a 90% level of confidence. What sample size should you gather to achieve a 0.45 hour margin of error? Round your answer up to the nearest whole number.

Answer 1

Answer:

The sample size should you gather to achieve a 0.45 hour margin of error

(n) = 411

Step-by-step explanation:

Step(i)

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)

Step(ii)

Margin of error = 1.69S / √n

Given data sample standard deviation S =5.4 hours.

margin of error = o.45

Margin of error   = $\frac{2S}{\sqrt{n} }$

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

Conclusion:-

The sample size should you gather to achieve a 0.45 hour margin of error

(n) = 411