Question

A physicist examines 10 water samples for iron concentration. The mean iron concentration for the sample data is 0.711 cc/cubic meter with a standard deviation of 0.0816. Determine the 90% confidence interval for the population mean iron concentration. Assume the population is approximately normal. Step 1 of 2 : Find the critical value that should be used in constructing the confidence interval. Round your answer to three decimal places.

Answer 1

Answer:

Critical value: T = 1.8331

The 90% confidence interval for the population mean iron concentration is between 0.561 cc/m³ and 0.861 cc/m³

Step-by-step explanation:

We have the standard deviation for the sample. So we use the t-distribution 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 = 10 - 1 = 9

90% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 9 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.9}{2} = 0.95$. So we have T = 1.8331, which is the critical value.

The margin of error is:

M = T*s = 1.8331*0.0816 = 0.15

In which s is the standard deviation of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 0.711 - 0.15 = 0.561 cc/m³

The upper end of the interval is the sample mean added to M. So it is 0.711 + 0.15 = 0.861 cc/m³

The 90% confidence interval for the population mean iron concentration is between 0.561 cc/m³ and 0.861 cc/m³

Answer 2

Answer:

$0.711 -1.833 \frac{0.0816}{\sqrt{10}}= 0.664$

$0.711 +1.833 \frac{0.0816}{\sqrt{10}}= 0.758$

So then we can conclude that the true mean at 95% of confidence is between $0.664 \leq \mu \leq 0.758$

Step-by-step explanation:

For this case we have the following info given:

$\bar X = 0.711$ represent the iron concentration for the sample selected

$s =0.0816$ represent the deviation for the sample

$n =10$ represent the sample size for the water samples

We want to find a confidence interval for the true ean and for this case the correct formula is given by:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$

We can find the critical value with the confidence level on this case is 0.9 or 90%. So then the significance level is $\alpha=1-0.9=0.1$ and the value of $\alpha/2 =0.05$.

The degrees of freedom for this special case are given by:

$df =n-1=10-1=9$

And the critical value using the t distribution with 9 degrees of freedom is given by:

$t_{\alpha/2}= 1.833$

And replacing into the formula for the confidence interval we got:

$0.711 -1.833 \frac{0.0816}{\sqrt{10}}= 0.664$

$0.711 +1.833 \frac{0.0816}{\sqrt{10}}= 0.758$

So then we can conclude that the true mean at 95% of confidence is between $0.664 \leq \mu \leq 0.758$