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

In summary, the conversation discusses a statistics question involving a table of pollution levels from 1950 to 1975. The significance test for unknown mean and standard deviation is used to determine if the change in pollution levels is significant. The formula for t is used to calculate a value of 2.47813, which is then looked up in a t-table at a 95% level of significance. The null hypothesis is that the change in pollution levels is 0, so the value of \mu_{0} is set to 0. The conversation also discusses finding the 95% confidence interval for the mean and variance, using the critical t and chi-squared values. Finally, the conversation ends with the person asking for corrections and
  • #1
Math_Frank
27
0

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.
 
Last edited:
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  • #3
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.
 
Last edited:
  • #4
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.
 

1. What is the significance level in statistics?

The significance level, denoted as α (alpha), is the probability of rejecting the null hypothesis when it is actually true. It is commonly set at 0.05 or 5%, meaning that there is a 5% chance of making a Type I error (rejecting the null hypothesis when it is true).

2. What is the confidence interval of the mean?

The confidence interval of the mean is a range of values that is likely to contain the true population mean with a certain level of confidence. It is calculated from a sample of data and is often used to estimate the population mean when the population standard deviation is unknown.

3. How is the confidence interval of the mean calculated?

The confidence interval of the mean is calculated using the sample mean and the standard error. The standard error takes into account the variability of the sample and is calculated by dividing the standard deviation by the square root of the sample size. The confidence interval can then be calculated as sample mean ± (critical value x standard error), where the critical value is determined by the desired level of confidence and the degrees of freedom.

4. What is the purpose of a confidence interval for the variance?

A confidence interval for the variance is used to estimate the true population variance with a certain level of confidence. It is calculated from a sample of data and can be used to assess the accuracy of a sample variance or to compare the variances of two populations.

5. How is the confidence interval of the variance calculated?

The confidence interval of the variance is calculated using the sample variance and the chi-square distribution. The upper and lower bounds of the confidence interval are determined by the degrees of freedom and the critical values from the chi-square distribution. The formula for calculating the confidence interval of the variance can be found in most statistics textbooks or online resources.

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