It costs you $10 to draw a sample of size n=1 and measure the attribute of interest. You have a budget of $1,200.

a. Do you have sufficient funds to estimate the population mean for the attribute of interest with a 95% confidence interval 4 units in width? Assume stdeviation(sigma) =12

b. If a 90& confidence level were used would your answer to part a change? explain.

It costs you $10 to draw a sample of size n=1 and measure the attribute of interest. You have a budget of $1,200.

a. Do you have sufficient funds to estimate the population mean for the attribute of interest with a 95% confidence interval 4 units in width? Assume stdeviation(sigma) =12

b. If a 90& confidence level were used would your answer to part a change? explain.

To answer part a, we need to calculate the required sample size to estimate the population mean with a 95% confidence interval of 4 units in width.

To calculate the required sample size, we can use the formula:

n = (Z * sigma / E)^2

Where:
n = required sample size
Z = Z-score corresponding to the desired confidence level (for 95% confidence level, Z = 1.96)
sigma = standard deviation of the population
E = margin of error (half the desired confidence interval width)

In this case, we want the confidence interval width to be 4 units. So the margin of error (E) is 4/2 = 2. Also, the given standard deviation is sigma = 12.

Plugging the values into the formula:

n = (1.96 * 12 / 2)^2
n = (23.52)^2
n ≈ 553

Therefore, we would need a sample size of approximately 553 to estimate the population mean with a 95% confidence interval of 4 units in width.

Now, to answer part b, if a 90% confidence level were used instead of 95%, the Z-score value would change. For a 90% confidence level, the Z-score is 1.645.

Using the same formula as before, but with the new Z-score:

n = (1.645 * 12 / 2)^2
n = (19.74)^2
n ≈ 389

Therefore, with a 90% confidence level and the same desired confidence interval width of 4 units, we would need a sample size of approximately 389.

In conclusion, for part a, since the required sample size is approximately 553 and your budget is $1,200, you have sufficient funds to estimate the population mean with a 95% confidence interval of 4 units in width.

For part b, even if the confidence level changes to 90%, the answer would remain the same as the required sample size is still less than the budgeted amount.