sample of 49 observations is taken from a normal population with a standard deviation of 10. The sample mean is 55. Determine the 99 percent confidence interval for the population mean. Give the upper limit.

Answer 1

To find the 99 percent confidence interval for the population mean, we need to use the formula for the confidence interval:

Confidence interval = sample mean ± (z-value * standard deviation / square root of sample size)

Since the sample size is 49 and the standard deviation is 10, we can substitute these values into the formula:

Confidence interval = 55 ± (z-value * 10 / √49)

To find the z-value corresponding to a 99 percent confidence level, we need to find the z-value that leaves 0.5 percent in each tail of the standard normal distribution. Since the confidence level is 99 percent, the remaining 1 percent is divided equally between the two tails, giving us 0.5 percent in each tail. Looking up the z-value for 0.5 percent in a standard normal distribution table, we find that it is approximately 2.576.

Substituting these values into the formula, we get:

Confidence interval = 55 ± (2.576 * 10 / √49)

Simplifying this equation:

Confidence interval = 55 ± (25.76 / 7)

Confidence interval = 55 ± 3.68

Therefore, the upper limit of the 99 percent confidence interval for the population mean is 55 + 3.68 = 58.68.