Level of significans and confidens interval of the mean and the variance

[Math_Frank]

Homework Statement

I am given the table

[tex]\begin{array}{ccccc} samplesite & 1950 & 1975 & Change \\ 1 & 7.30 & 7.70 & 0.40\\ 2 & 6.14 & 6.72 & 0.58 \\ 3 & 6.47 & 6.32 & -0.15 \\4 & 5.87 & 6.38 & 0.51 \\ 5 & 6.06 & 6.34 & 0.28 \\ 6 & 4.71 & 5.78 & 1.07 \\ 7 & 5.45 & 5.59 & 0.14 \\ 8 & 6.17 & 6.33 & 0.16 \\9 & 5.83 & 5.50 & -0.33 \\ 10 & 5.55 & 5.55 & 0.00 \\ 11 & 5.50 & 6.08 & 0.58 \end{array}[/tex]

I have to stachistics question where the above table is used.

Homework Statement

Question (1)

Look at the change column: Is the change in polution levels from 1950 to 1975 significant?

Using the significance test for unknown mean and unknown standard diviation.

So I use the formula:

[tex]t = \frac{\overline{x} - \mu_0}{\frac{s}{\sqrt{n}}}[/tex]

where [tex]\overline{x} = \frac{\sum(x_i)}{N} = \frac{3.24}{11} = 0.2946[/tex]

and where [tex]S^2 = \frac{1}{N-1} \sum(X -\overline{x})^2 = 0.154[/tex]

where S = [tex]\sqrt{S^2} = 0.394228[/tex]

Average change = average 1975 - average 1950. Under the null hypothesis, change = 0, then mu = 0

Then I insert into the formula
[tex]t = \frac{0.2946 - 0}{\frac{0.392428}{\sqrt{11}}} = 2.47813.[/tex]

Looking that up in the t-table Level of significans is 95%. How does that sound?

Homework Statement

Homework Equations

But what do I use for [tex]\mu_{0}[/tex] ??

Question (2):

Write the 95% confidence Interval for the mean and the variance for the change polution from 1950 og 1975.

The 95% confidence interval for the mean is defined

[tex]\overline{x} + t_{CL} \cdot {N} , \overline{x} - t_{CL} \cdot {N} [/tex]

I know that [tex]\overline{x} = 0.2945[/tex] and if look at the t table for 95% t-level at degree of freedom df = N - 1 then I find a t-value called t = 2.98.

Therefore (as I understand it) the 95% confidence level for the mean is
[tex]0.2945 + 2.95 \cdot {11} , 0.2945 - 2.98 \cdot {11} [/tex] ?

If the above is true how do I write the 95% Confidance Interval of the variance?

Sincerely Yours
Math Frank.

 

[EnumaElish]

Average change = average 1975 - average 1950. Under the null hypothesis, change = 0 so [itex]\mu_0[/itex] = 0.

For (2) see this thread: https://www.physicsforums.com/showthread.php?t=214872&highlight=significance

 

[Math_Frank]

Hi I have changed the answer to question 1) With Your corrections.

Regarding question 2)

Is my assumption regarding the 95% interval for the mean. Is that correct?

If not I would very much appriacte if you could find it in your heart to correct 2) for me because, tomorrow I am going in for a bypass(believe it or not, I have had pain in my chest for some time now :( ), and I have to hand this in before I go to hospital. I know it much to ask, but it will only be this one time.

Sincerely Yours
Frank.

 

[EnumaElish]

The 95% C.I. around the mean is [itex]\bar {x} \pm t_{0.025}(n-1)s/\sqrt{n}[/itex] where [itex]\bar {x}[/itex] and s are respectively the computed average and the computed standard deviation of the sample, n is the sample size and [itex]t_{0.025}(n-1)[/itex] is the critical t value with 0.05 total (two-sided) tail probability and n-1 degrees of freedom.

The 95% C.I. around the variance is [L , U] = [ [itex](n-1)s^2\left/\chi^2_{0.025}(n-1)[/itex] , [itex](n-1)s^2\left/\chi^2_{0.975}(n-1)[/itex] ], where [itex]\chi^2_p(n-1)[/itex] is the critical chi-squared value with probability = p (to the left of the critical value) and degrees of freedom = n-1.

Good luck with the bypass.