Can someone answer the last two questions
2 answers:
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Answer:
The margin of error for a 95% confidence interval estimate for the population mean using the Student's t-distribution is of 6.22 ounces.
Step-by-step explanation:
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
95% 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 = 2.0739
The margin of error is:
M = T*s
In which s is the standard deviation of the sample.
In this question:
s = 3.
Then
M = 2.0739*3 = 6.22
The margin of error for a 95% confidence interval estimate for the population mean using the Student's t-distribution is of 6.22 ounces.