Question

1. Express the confidence interval (35%,42%) in the form of p±ME.

% ± %

2.

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=70 hours with a standard deviation of s=4.2 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 98% 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.

n = bacteria

If you can kindly answer both questions please I would really appreciate it. I have posted these questions several times and keep getting the wrong answers. Thank you so much

Answer 1

Question 1 :

p = sample proportion = p and ME = margin of error

P : Population proportion.

The confidence interval for population proportion is given by

(35%, 42%)

Upper confidence limit = p- M.E = 0.35 --------------(II)

Lower confidence limit = p + ME = 0.42 -------------(II)

From (I) & (II)

2p = 0.35 + 0.42

p = 0.385

Margin of error is given by

ME = 0.42 - 0.385 = 0.035

Required Answer : 38.5% 3.5%.

Question 2:

Consider

X : Lifespan of certain species of bacteria.

From the information

: Sample mean lifespan for species of bacteria.

S: sample standard deviation .

: mean lifespan for species of bacteria

98% confidence interval for mean lifespan for species of bacteria is

Margin of error is given by

Alpha = level of significance = 0.02

from normal table

Given: ME = 0.45

Hence required sample size is given by

n = 471.45

But n must be an integer

Required sample size = n = 471.