a random sample of 42 college graduates who worked during their program revealed that a student spent an averge of 5.5 yeas on the job before being promoted. The sample standard deviation was 1.1 yrs. Using the .99 degree of confidence, what is the confidence interval for the population mean
choices 5.04 and 5.96, 5.06 and 5.94, 2.67 and 8.33 4.4 and 6.60
5.06 and 5.94
5.04 and 5.96 is the correct answer
To find the confidence interval for the population mean, we can use the formula:
Confidence Interval = sample mean ± (critical value × standard deviation / square root of sample size)
In this case, the sample mean is 5.5 years, the sample standard deviation is 1.1 years, and the sample size is 42.
First, we need to find the critical value. Since the confidence level is .99, the significance level is 1 - .99 = 0.01. Since this is a two-tailed test, we divide the significance level by 2, giving us 0.01 / 2 = 0.005.
We can then look up the critical value associated with a 0.005 significance level in a t-distribution table with (n-1) degrees of freedom, where n is the sample size. In this case, since the sample size is 42, the degrees of freedom would be 42 - 1 = 41.
Using the t-distribution table or a statistical calculator, we find that the critical value for a 0.005 significance level with 41 degrees of freedom is approximately 2.704.
Plugging these values into the formula, the confidence interval is:
Confidence Interval = 5.5 ± (2.704 × 1.1 / √42)
Calculating this, we get:
Confidence Interval = 5.5 ± (2.704 × 0.1698)
Confidence Interval = 5.5 ± 0.4593
This gives us a confidence interval of approximately (5.041, 5.959).
Therefore, the correct choice from the given options is 5.04 and 5.96.
To calculate the confidence interval for the population mean, we can use the following formula:
Confidence interval = sample mean ± (critical value * sample standard deviation / square root of the sample size)
In this case, the given information is as follows:
Sample size (n) = 42
Sample mean (x̄) = 5.5 years
Sample standard deviation (σ) = 1.1 years
Since we want a 99% confidence level, we need to find the critical value associated with the appropriate level of significance (α). For a 99% confidence level, the alpha level is 1 - 0.99 = 0.01.
To find the critical value, you can consult a t-distribution table or use statistical software. For simplicity, let's assume that the critical value is approximately 2.6 based on normal approximation.
Substituting the values into the formula, we get:
Confidence interval = 5.5 ± (2.6 * 1.1 / √42)
Confidence interval = 5.5 ± 0.421
Calculating this, we have two extreme values for the confidence interval:
Lower bound = 5.5 - 0.421 = 5.079
Upper bound = 5.5 + 0.421 = 5.921
Therefore, the confidence interval for the population mean is approximately 5.08 to 5.92.
Among the given choices, the correct answer is 5.06 and 5.94.