Can someone help me with these two?

Question

You are interested in estimating the the mean weight of the local adult population of female white-tailed deer (doe). From past data, you estimate that the standard deviation of all adult female white-tailed deer in this region to be 23 pounds. What sample size would you need to in order to estimate the mean weight of all female white-tailed deer, with a 96% confidence level, to within 7 pounds of the actual weight?

 

You are a researcher studying the lifespan of a certain species of bacteria. A preliminary sample of 25 bacteria reveals a sample mean of `\bar{x}=76` 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 0.8 hours at a 90% level of confidence.

What sample size should you gather to achieve a 0.8 hour margin of error? Round your answer up to the nearest whole number.

n = __________________ bacteria

Answer 1

UPDATED SOLUTION:

To determine the sample size needed to estimate the mean weight of all female white-tailed deer with a certain level of confidence and margin of error, we can use the formula:

n = (Z * σ / E)^2

Where:

n = sample size

Z = Z-score corresponding to the desired confidence level

(in this case, for a 96% confidence level, Z ≈ 2.053)

σ = standard deviation of the population

E = margin of error (desired maximum deviation from the mean weight)

Plugging in the values:

Z ≈ 2.053

σ = 23 pounds

E = 7 pounds

n = (2.053 * 23 / 7)^2

Calculating this expression:

n=45.502733898

Rounding up to the nearest whole number, the sample size needed would be approximately 46

To determine the sample size needed to estimate the mean lifespan of the bacteria species with a certain margin of error at a specific confidence level, we can use the formula:

n = (Z * σ / E)^2

Where:

n = sample size

Z = Z-score corresponding to the desired confidence level (for a 90% confidence level, Z ≈ 1.645)

σ = standard deviation of the population (unknown in this case, but we can estimate it using the sample standard deviation)

E = margin of error (0.8 hours)

Given:

Z ≈ 1.645

s = 4.2 hours

E = 0.8 hours

Using the formula, we can calculate the sample size:

n = (1.645 * 4.2 / 0.8)^2

n = 74.5848140625

Rounding up to the nearest whole number, the sample size needed is 75.