An investigative question in your employee satisfaction study seeks to assess employee ""job involvement."" Create a measurement
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Answer:
The 95% confidence interval for the mean GPA of all accounting students at this university is between 2.5851 and 3.2549
Step-by-step explanation:
We are in posessions of the sample's standard deviation. So we use the student's t-distribution to solve this question.
The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So
df = 20 - 1 = 19
95% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 19 degrees of freedom(y-axis) and a confidence level of ). So we have T = 2.0930
The margin of error is:
M = T*s = 2.0930*0.16 = 0.3349.
In which s is the standard deviation of the sample.
The lower end of the interval is the sample mean subtracted by M. So it is 2.92 - 0.3349 = 2.5851
The upper end of the interval is the sample mean added to M. So it is 2.92 + 0.3349 = 3.2549
The 95% confidence interval for the mean GPA of all accounting students at this university is between 2.5851 and 3.2549