Number 30 does anybody know the answer to this
1 answer:
You might be interested in
Answer:
The 88% confidence level for the average weight gain if between -1.30 lbs and 4.58 lbs.
Step-by-step explanation:
We have the standard deviation for the sample, which means that the t-distribution is used 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 = 23 - 1 = 22
88% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 22 degrees of freedom(y-axis) and a confidence level of . So we have T = 1.6176
The margin of error is:
In which s is the standard deviation of the sample and n is the size of the sample.
The lower end of the interval is the sample mean subtracted by M. So it is 1.64 - 2.94 = -1.30 lbs
The upper end of the interval is the sample mean added to M. So it is 1.64 + 2.94 = 4.58 lbs
The 88% confidence level for the average weight gain if between -1.30 lbs and 4.58 lbs.