A random sample of 100 students from a high school was selected. If a 98% confidence interval estimate for the mean age of students is calculated to be 14.2511 to 14.9489, then the standard deviation of the sample mean is?
2E + 14.2511 = 14.9489
2E = 0.6978
E = 0.3489
E = za/2 * s /sqrt(n))
.3489 = 2.33s/10
. 3489 = .233s
s = 1.49742
Standard deviation. = 1.49742
To calculate the standard deviation of the sample mean, we need to know the formula for a confidence interval and the sample size.
The formula for a confidence interval for the mean is:
CI = X̄ ± Z*(σ/√n),
where CI is the confidence interval, X̄ is the sample mean, Z is the z-score corresponding to the desired confidence level, σ is the population standard deviation (which we want to find), and n is the sample size.
In this case, the confidence level is 98%, so the corresponding Z-value can be found using a z-table or a statistical calculator. For a 98% confidence interval, the Z-value is approximately 2.33.
Now let's use the formula to solve for σ:
CI = 14.2511 to 14.9489
X̄ = (14.2511 + 14.9489) / 2 = 14.6 (mean of the confidence interval)
Z = 2.33 (Z-value for 98% confidence level)
n = 100 (sample size)
CI = X̄ ± Z*(σ/√n)
14.6 = (14.2511 + 14.9489) / 2 = 14.6 ± 2.33*(σ/√100)
Now we can rearrange the equation to solve for σ:
2.33*(σ/√100) = 14.6 - 14.6
2.33*(σ/10) = 0
σ/10 = 0 / 2.33
σ/10 = 0
Therefore, the standard deviation of the sample mean is 0.