Isaac Barry

2022-09-29

The confidence interval is [2.663;2.937]?

The concentration of calcium in the blood for a given population follows a normal law of mean $\mu =2.8$ mmol / L and standard deviation $\sigma =0.7$ mmol / L. If we take a sample of 100 people, what is the confidence interval of the calcium concentration at the 95% confidence level.

I got the confidence interval [2.663; 2.937] but I think this is wrong.

The concentration of calcium in the blood for a given population follows a normal law of mean $\mu =2.8$ mmol / L and standard deviation $\sigma =0.7$ mmol / L. If we take a sample of 100 people, what is the confidence interval of the calcium concentration at the 95% confidence level.

I got the confidence interval [2.663; 2.937] but I think this is wrong.

oldgaffer1b

Beginner2022-09-30Added 9 answers

Step 1

We know the population standard deviation $\sigma =0.7$ so we can find z∗ by:

$${z}^{\ast}={\mathrm{\Phi}}^{-1}(1-\frac{\alpha}{2})$$

Step 2

As we want a 95% confidence level we have $\alpha =0.05$, this means ${z}^{\ast}=1.96$. We also know the population mean $\mu =2.8$ so we can get the confidence interval by plugging in the values (including $n=100$):

$$(\mu -{z}^{\ast}\frac{\sigma}{\sqrt{n}},\mu +{z}^{\ast}\frac{\sigma}{\sqrt{n}})=(2.6628,2.9372)$$

We know the population standard deviation $\sigma =0.7$ so we can find z∗ by:

$${z}^{\ast}={\mathrm{\Phi}}^{-1}(1-\frac{\alpha}{2})$$

Step 2

As we want a 95% confidence level we have $\alpha =0.05$, this means ${z}^{\ast}=1.96$. We also know the population mean $\mu =2.8$ so we can get the confidence interval by plugging in the values (including $n=100$):

$$(\mu -{z}^{\ast}\frac{\sigma}{\sqrt{n}},\mu +{z}^{\ast}\frac{\sigma}{\sqrt{n}})=(2.6628,2.9372)$$

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