Is 5,690 divisible by 10

The 98% confidence interval for the mean magnesium ion concentration is between 122 ppm and 220 ppm. As the lower bound of the interval is below 195, it is not reasonable to conclude that the mean magnesium ion concentration may be greater than 195

Step-by-step explanation:

The six measurements are:

202, 164, 157, 177, 113, 213

Sample mean:

Sum of all values divided by the number of values which is 6. So

$S_m = \frac{202+164+157+177+113+213}{6} = 171$

Sample standard deviation:

Square root of the difference squared between each value and the mean, and divided by 1 subtracted by the sample size. So

$s = \sqrt{\frac{(202-171)^2+(164-171)^2+(157-171)^2+(177-171)^2+(113-171)^2+(213-171)^2}{5}} = 35.69$

Confidence interval

We have the standard deviation for the sample, which means that the t-distribution will be 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 = 6 - 1 = 5

98% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 5 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.98}{2} = 0.99$ . So we have T = 3.365

The margin of error is:

$M = T\frac{s}{\sqrt{n}} = 3.365\frac{35.69}{\sqrt{6}} = 49$

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 171 - 49 = 122 ppm

The upper end of the interval is the sample mean added to M. So it is 171 + 49 = 220 ppm

The 98% confidence interval for the mean magnesium ion concentration is between 122 ppm and 220 ppm. As the lower bound of the interval is below 195, it is not reasonable to conclude that the mean magnesium ion concentration may be greater than 195

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3 years ago