a study found that the average time it took a person to find a new job was 130 days. If a sample of 36 job seekers was surveyed, find the 95% confidence interval of the mean. Assume the standard deviation of the sample is 12 days.
CI95 = mean + or - 1.96(sd divided by √n)
...where + or - 1.96 represents the 95% confidence interval using a z-table, sd = standard deviation, √ = square root, and n = sample size.
Using your data:
CI95 = 130 + or - 1.96 (12/√36)
I'll let you finish the calculation to determine the interval.
I hope this will help.
1.8 3.2
To find the 95% confidence interval of the mean, we can use the formula:
Confidence Interval = sample mean ± (z-value * standard deviation / √n)
In this case, we are given the following information:
- Sample mean: 130 days
- Sample size: 36 job seekers
- Standard deviation: 12 days
The z-value for a 95% confidence interval is 1.96. Plug these values into the formula to calculate the confidence interval:
Confidence Interval = 130 ± (1.96 * 12 / √36)
First, we calculate the standard error of the mean, which is the standard deviation divided by the square root of the sample size:
Standard Error = 12 / √36 = 12 / 6 = 2
Now, plug in the values to calculate the confidence interval:
Confidence Interval = 130 ± (1.96 * 2) = 130 ± 3.92
The lower boundary of the confidence interval is the sample mean minus the margin of error:
Lower Boundary = 130 - 3.92 = 126.08
The upper boundary of the confidence interval is the sample mean plus the margin of error:
Upper Boundary = 130 + 3.92 = 133.92
Therefore, the 95% confidence interval of the mean is (126.08, 133.92).