Answer:
The 99% confidence interval for the true mean checking account balance for local customers is ($439.29, $888.99).
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 = 14 - 1 = 13
99% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 13 degrees of freedom(y-axis) and a confidence level of
. So we have T = 3.0123
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 664.14 - 224.85 = $439.29
The upper end of the interval is the sample mean added to M. So it is 664.14 + 224.85 = $888.99.
The 99% confidence interval for the true mean checking account balance for local customers is ($439.29, $888.99).
We don’t know the number of employees remaining at the company.
We don’t know the total number of employees remaining at the company.
The classification consists of gender, division, and employment type.
It doesn’t matter what order those attributes come in. A person who is male, age group 6, and employment type 10 has the same classification as someone who is of employment type 6, male, and age group10.
Therefore since order does not matter it is a combination question.
Assuming only two genders are allowed (you didn’t say and some people think there are about 10 of them)
It is either 120 (2 times 6 times 10) or "None of the above" if you believe there are more than 2 genders.
2×6×10=120
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The product of 100 x a number would be 10 times larger than the product of 11 x a number