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choli [55]
3 years ago
13

Anyone know the answer??

Mathematics
1 answer:
Xelga [282]3 years ago
3 0

Answer

3.5 x 10^{3}

Step-by-step explanation

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A news report states that the 99% confidence interval for the mean number of daily calories consumed by participants in a medica
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Answer:

The sample mean is of 1925 calories.

The margin of error is of 75 calories.

The sample standard deviation is of 109.7992 calories.

Step-by-step explanation:

Sample mean:

The sample mean is the mean value of the two bounds of the confidence interval. So

M = \frac{1850 + 2000}{2} = 1925

The sample mean is of 1925 calories.

The margin of error

Difference between the bounds and the sample mean. So

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The margin of error is of 75 calories.

Sample standard deviation:

Here, I am going to expand on the t-distribution.

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 = 18 - 1 = 17

99% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 17 degrees of freedom(y-axis) and a confidence level of 1 - \frac{1 - 0.99}{2} = 0.995. So we have T = 2.898

The margin of error is:

M = T\frac{s}{\sqrt{n}}

In which s is the standard deviation of the sample and n is the size of the sample.

Since M = 75, T = 2.898, n = 18

M = T\frac{s}{\sqrt{n}}

75 = 2.898\frac{s}{\sqrt{18}}

s = \frac{75\sqrt{18}}{2.898}

s = 109.7992

The sample standard deviation is of 109.7992 calories.

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