Mathematics: Hannibal wants to estimate the true average brain mass, u, but he is locked in a cell with no internet and only has a scale. He collects a small random sample of size n = 5 from nearby prison guards. He find that these brains have the following weights: 1320g, 1400g, 1300g, 1460g, 1350g Suppose that human brains follow a normal distribution, and that Hannibal s sample is representative of the entire population. a. (1 pt) Compute the sample standard deviation, s, of these brains. Do not use a computer

yeetStep-by-step explanation:urmom...

Hannibal wants to estimate the true average brain

Ответ:

claytonhopkins

23.01.2022

Mathematics

a) s=64.65

b) $1309.33\leq \mu \leq 1422.67$

c) $1327.26\leq \mu \leq 1404.74$

d) $\mu \geq 1318.44$

Step-by-step explanation:

a) To calculate the sample standard deviation, first we have to calculate the sample mean.

$\bar x=\frac{1}{n} \sum x_i=(1/5)*(1320+1400+1300+1460+1350)\\\\\bar x=(1/5)*6830=1366$

Now, the standard deviation is:

$s=\sqrt{\frac{1}{n-1}\sum (x_i-\bar x)^2}$

$\sqrt{\frac{1}{5-1}*[(1320-1366)^2+(1400-1366)^2+(1300-1366)^2+(1460-1366)^2+(1350-1366)^2]}$

$s=\sqrt{\frac{1}{4} [(-46)^2+(34)^2+(-66)^2+(94)^2+(-16)^2]}\\\\\\s=\sqrt{\frac{1}{4}[2116+1156+4356+8836+256]}\\\\s=\sqrt{\frac{1}{4}*16720}\\\\s=\sqrt{4180}\\\\s=64.65$

b) The z-value for a 95% confidence interval is z=1.96.

The margin of error is

$E=z\sigma/\sqrt{n}=1.96*64.65/\sqrt{5}=126.714/2.236=56.67$

The lower and upper limit of the CI are:

$LL=\bar x-z\sigma/\sqrt{n}=1366-56.67=1309.33\\\\UL=\bar x+z\sigma/\sqrt{n}=1366+56.67=1422.67$

The confidence interval is:

$1309.33\leq \mu \leq 1422.67$

c) In this case, the z-value for a 82% CI is z=1.34.

The margin of error is

$E=z\sigma/\sqrt{n}=1.34*64.65/\sqrt{5}=86.631/2.236=38.74$

The lower and upper limit of the CI are:

$LL=\bar x-z\sigma/\sqrt{n}=1366-38.74=1327.26\\\\UL=\bar x+z\sigma/\sqrt{n}=1366+38.74=1404.74$

The confidence interval is:

$1327.26\leq \mu \leq 1404.74$

c) Now we have to construct a one sided 95% confidence interval, with only one bound (lower bound).

The z-value is z=1.645

The margin of error is

$E=z\sigma/\sqrt{n}=1.645*64.65/\sqrt{5}=106.35/2.236=47.56$

The lower limit of the CI is:

$LL=\bar x-z\sigma/\sqrt{n}=1366-47.56=1318.44$

The confidence interval is:

$\mu \geq 1318.44$

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