How large a sample should be taken if the population mean is to be estimated with 99% confidence to within $72? The population has a standard deviation of $800.
mu = bar x +/- 2.58(sigma/sqrt N)
mu is the population mean.
bar x is the average.
2.59 gives the 99% confidence interval.
sigma is standard deviation. N is the number of sample. Solve for N.
Well, if you want to estimate the population mean with 99% confidence to within $72, I would suggest taking a sample of zero. Who needs data when you have confidence, am I right? Just kidding! Let me give you a serious answer.
To determine the sample size, we can use the formula:
n = (Z * σ / E)^2
where:
n = sample size
Z = Z-score for the desired confidence level (99% confidence level corresponds to a Z-score of approximately 2.576)
σ = standard deviation of the population
E = margin of error (which is given as $72)
By plugging in the values into the formula, we get:
n = (2.576 * 800 / 72)^2
n ≈ 367.7
So, you should round up and take a sample size of at least 368 to estimate the population mean with 99% confidence and a margin of error of $72. Just make sure your clown car is big enough to fit all those samples in!
To determine the sample size needed to estimate the population mean with a certain level of confidence, you can use the formula:
n = (Z * σ / E)²
Where:
n = sample size
Z = Z-score corresponding to the desired level of confidence
σ = standard deviation of the population
E = margin of error (half the desired confidence interval)
In this case, we want a 99% confidence level, which corresponds to a Z-score of 2.576 (obtained from a standard normal distribution table). The margin of error, E, is given as $72.
Plugging in the values:
n = (2.576 * 800 / 72)²
n ≈ (2060.8 / 72)²
n ≈ 28.6²
n ≈ 821.16
Therefore, a sample size of approximately 821 would be needed to estimate the population mean with 99% confidence to within $72. Since you cannot have a fraction of a sample, round up to the nearest whole number, giving a final sample size of 822.
To determine the sample size needed to estimate the population mean with a certain level of confidence, we can use the formula:
n = (Z * σ / E)^2
Where:
- n is the required sample size
- Z is the z-score corresponding to the confidence level (in this case, for a 99% confidence level, Z would be approximately 2.58)
- σ is the population standard deviation
- E is the desired margin of error (in this case, $72)
Plugging in the given values, we can compute the necessary sample size:
n = (2.58 * 800 / 72)^2
n = 28.5^2
n ≈ 812.25
Therefore, a sample size of approximately 812 should be taken to estimate the population mean with 99% confidence and a margin of error of $72. Since you cannot have a fraction of a person in your sample, we would round up this value to the nearest integer, making the final answer a sample size of 813.