Please help me about this math
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Answer:
The 98% confidence interval for the mean usage in the March quarter of 2006, in kWh, was (333.87, 416.13).
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 = 30 - 1 = 29
98% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 29 degrees of freedom(y-axis) and a confidence level of . So we have T = 2.462
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 375 - 41.13 = 333.87 kWh
The upper end of the interval is the sample mean added to M. So it is 375 + 41.13 = 416.13 kWh
The 98% confidence interval for the mean usage in the March quarter of 2006, in kWh, was (333.87, 416.13).