suppose you wish to estimate a population mean based on a random sample of n observations, and prior experience suggests that sd=12.7. if you wish to estimate the mean correct to within 1.6, with probability equal to .95, how many observations should be included in your sample?
Formula:
n = [(z-value * sd)/E]^2
...where n = sample size, z-value will be 1.96 using a z-table to represent the 95% confidence interval, sd = 12.7, E = 1.6, ^2 means squared, and * means to multiply.
I'll let you take it from here.
To determine the minimum sample size needed to estimate the population mean, we can use the formula for sample size calculation:
n = (Z * σ / E) ^ 2
Where:
- n is the required sample size
- Z is the z-score corresponding to the desired level of confidence (in this case 0.95)
- σ is the known standard deviation of the population (12.7 in this case)
- E is the desired margin of error (1.6 in this case)
First, we need to find the z-score for a confidence level of 0.95. This can be found using a standard normal distribution table or using statistical software. For a 95% confidence level, the z-score is approximately 1.96.
Now, we can substitute the values into the formula:
n = (1.96 * 12.7 / 1.6) ^ 2
Calculating this expression:
n = (22.792 / 1.6) ^ 2
n = 14.2455 ^ 2
n ≈ 203
Therefore, you should include at least 203 observations in your sample to estimate the population mean within 1.6 with a 95% confidence level.
To estimate the sample size needed to estimate a population mean with a desired level of accuracy, we can use the formula for the confidence interval:
n = (Z * σ / E)²
Where:
n is the required sample size
Z is the desired level of confidence, which corresponds to the Z-value for the desired confidence level (e.g., Z=1.96 for a 95% confidence level)
σ is the estimated standard deviation of the population
E is the desired margin of error
In this case, you want to estimate the mean correct to within 1.6 units with a probability of 0.95, and prior experience suggests that the standard deviation (σ) of the population is 12.7.
First, let's find the Z-value corresponding to a 95% confidence level. You can use a Z-table or a statistical software to find that the Z-value for a 95% confidence level is approximately 1.96.
Now we can substitute the values into the formula:
n = (1.96 * 12.7 / 1.6)²
n = (19.84 / 1.6)²
n ≈ (12.4)²
n ≈ 153.76
Since you cannot have a fractional sample size, you would need at least 154 observations in your sample to estimate the population mean with a margin of error of 1.6 units and a confidence level of 95%.