Twenty-five blood samples were selected by taking every seventh blood sample from racks holding 187 blood samples from the morning draw at a medical center. The white blood count (WBC) was measured using a Coulter Counter Model S. The mean WBC was 8.636 with a standard deviation of 3.9265.
(a) Construct a 90 percent confidence interval for the true mean.
(b) Why might
normality be an issue here?
(c) What sample size would be needed for an error of ± 1.5 with
98 percent confidence?
(a) To construct a 90 percent confidence interval for the true mean, we can use the formula:
Confidence Interval = sample mean ± critical value * (standard deviation / sqrt(sample size))
In this case, the sample mean is 8.636, the standard deviation is 3.9265, and the sample size is 25.
First, we need to find the critical value. Since we want a 90 percent confidence level, we need to find the z-value that corresponds to a 95 percent confidence level (since we're dealing with a two-tailed distribution).
Using a standard normal distribution table or calculator, the z-value for a 95 percent confidence level is approximately 1.645.
Therefore, the confidence interval is:
Confidence Interval = 8.636 ± 1.645 * (3.9265 / sqrt(25))
Simplifying the equation:
Confidence Interval = 8.636 ± 1.645 * 0.7853
The lower limit of the confidence interval is:
8.636 - 1.645 * 0.7853 = 8.636 - 1.291 = 7.345
The upper limit of the confidence interval is:
8.636 + 1.645 * 0.7853 = 8.636 + 1.291 = 9.927
So, the 90 percent confidence interval for the true mean is (7.345, 9.927).
(b) Normality might be an issue here because in order to perform certain statistical tests and construct confidence intervals, the sample data needs to follow a normal distribution. If the data is not normally distributed, the results may be inaccurate or unreliable.
In this case, the sample size is relatively small (only 25), so it may be difficult to determine whether the underlying population distribution is truly normal. Additionally, the measurement of the white blood count (WBC) might not necessarily follow a perfect normal distribution.
It's important to note that violating the assumption of normality doesn't necessarily mean that the confidence interval is completely invalid, but it may affect the accuracy and reliability of the results.
(c) To determine the sample size required for an error of ±1.5 with 98 percent confidence, we can use the formula:
Required Sample Size = (Z-value * standard deviation / desired margin of error)²
In this case, the desired margin of error is ±1.5, which means the margin of error would be 3 (2*1.5) for a two-tailed confidence interval.
We need to find the Z-value that corresponds to a 99 percent confidence level since we're dealing with a two-tailed distribution and want to achieve 98 percent confidence.
Using a standard normal distribution table or calculator, the Z-value for a 99 percent confidence level is approximately 2.576.
Therefore, the required sample size is:
Required Sample Size = (2.576 * 3.9265 / 3)²
Simplifying the equation:
Required Sample Size = (2.576 * 1.3088)²
Required Sample Size = 8.36²
Required Sample Size ≈ 69.77
Thus, a sample size of approximately 70 would be needed for an error of ±1.5 with 98 percent confidence.