An investment adviser states that the risk (standard deviation) when investing in a certain instrument is greater than $3.$ To corroborate the validity of this statement, he plans to realize: $H_0: \sigma=3$ vs. $H_1: \sigma>3$ with an aleatory sample of $n = 29$ performance values.

Then he gets ​a sample mean of $\bar X =8$ and a sample variance of $S^2 = 16.$ What is the approximate value of the significance level of the test if the rejection region (critical region) given by $S^2$ is such that $S^2 \ge 13?$

Note: $\sigma$ denotes population standard deviation and $S^2$ denotes sample variance.

My procedure: I looked up for the value 13 with 28 (29-1) degrees of freedom in the Chi squared table to know the significance level, but that did not work... The answer is supposed to be: $.025<$ p-value $<.05.$

  • $\begingroup$ Use $\frac{(n-1)S^2}{\sigma^2} \sim \mathsf{Chisq}(n-1)$ to find significance level $\alpha = P(S^2 > 13 | \sigma = 3).$ // You say your sample has $\sigma = 4,$ but you're testing $H_0: \sigma = 3$ vs. $H_1: \sigma > 3.$ In order to find the significance level of the test you need to use the 'null distribution' with $\sigma = 3.$ You might do a power computation for the alternative $\sigma_a = 4.$ Please edit your question to clarify. $\endgroup$ – BruceET Dec 14 '18 at 18:45
  • $\begingroup$ Thanks for the clarifications. I edited a little more to use math symbols and punctuation more clearly. Please look to see if I have changed your meaning. // I am also making some corresponding changes to my answer. $\endgroup$ – BruceET Dec 14 '18 at 21:43
  • $\begingroup$ You´re right Bruce, I looked up for the values in the table and the p-value is between those values (.5% < p-value < 1%). Thanks for your complete answer. $\endgroup$ – Luis Dec 15 '18 at 4:03
  • $\begingroup$ OK, now that we agree on the Question and the Answer, I hope you will take the time to make sure you understand all of the steps in between. Time to solidify the connection between normal and chi-squared distributions; time to make sure you know how to use chi-squared tables, time to memorize definitions of significance level and p-value. $\endgroup$ – BruceET Dec 15 '18 at 18:49

Comment continued. I assume you are using normal data. It is not possible to work such a problem without some assumption about the population distribution. At an elementary level, the background should have been covered to allow answering for normal data. For example, the assumption of normal data leads directly to the chi-squared distribution of $Q$ below.

The significance level is the probability of rejection $P(S^2 \ge 13)$ under the assumption that $\sigma = 3.$ I guess your assignment is to get an answer using the relationship in my Comment above: With $Q = \frac{(n-1)S^2}{\sigma^2},$ you have $Q \sim \mathsf{Chisq}(\nu),$ where $\nu = n-1 = 29-1 = 28$ degrees of freedom.

For your data with $n = 29, S^2 = 16$ and your null hypothesis $\sigma = 3,$ you have $$Q = \frac{(n-1)S^2}{\sigma^2} = \frac{28(16)}{9} = 49.78.$$

Now, the critical value for a test is given as $S^2 > 13.$ But the printed chi-squared table is given in terms of $Q,$ which is the chi-squared random variable. So $S^2 > 13$ is equivalent to $$Q_c = \frac{28S^2}{9} = \frac{28(13)}{9} = 40.44.$$

So you reject $H_0,$ according to the stated critical region, either because $S^2 = 16 > 13$ or because $Q = 49.78 > 40.44.$

If you look at the row for DF = 28 in your printed chi-squared table, you will find that 49.78 lies between the entries 48.27 and 50.99, in columns headed .01 and .005., indicating that the values cut about 1% and 0.5% from the upper tail of the relevant chi-squared distribution. So I get that the p-value lies between 1% and 0.5%, not between 5% and 0.25% as in the answer provided. The exact p-value is 0.007, which lies between 0.01 and 0.005.

qchisq(c(.9, .95, .975, .99, .995), 28)
[1] 37.91592 41.33714 44.46079 48.27824 50.99338
1 - pchisq(49.78, 28) 
[1] 0.006842861

The critical values implied by $S^2 \ge 13$ or $Q > 40.44$ are not for a test at the 5% significance level. The level based on those critical values is about 6%.

1 - pchisq(40.44, 28)
[1] 0.06038755

A test at the 5% level would use $Q \ge 41.34$ or $S^2 = 9(41.34)/28 \approx 13.29.$

For verification, a simple simulation in R gets approximately these same results (6%, 13.3, 0.007), directly from the simulated distribution of $S^2,$ when $n = 29$ and $\sigma = 3.$

set.seed(1214);  m = 10^5;  n = 29;  sg = 3;  mu = 8
v = replicate(m, var(rnorm(n, mu, sg)))
mean(v > 13)
[1] 0.06023
quantile(v, .95)
mean(v > 16)
[1] 0.00683

The figure below shows a histogram of the simulated values of $Q$ with the density function of $\mathsf{Chisq}(28).$ The heavy vertical black line is at the observed value of $Q$ corresponding to $S^2 = 16.$ Thin vertical brown and cyan lines cut probabilities 5% and 1%, respectively, from the upper tail of the distribution.

enter image description here

So I don't get the alleged answer. I don't know whether we are dealing with a misinterpretation of the problem, with typgraphical/computational errors, or with an incorrect answer. Please check everything.

Notes: (1) In R statistical software the function pchisq denotes the CDF of a chi-squared distribution and the function qchisq indicates its inverse CDF or 'quantile' function. (2) In the simulation v is a vector of 100,000 simulated values of $S^2.$

  • $\begingroup$ Thanks, BruceET. I have already edited the question. But why did you use a normal distribution if , in general, a chi-squared is used for variances? Anyways, I also put the answer of my teacher below. Hope you can help me to figure out why is that the answer. $\endgroup$ – Luis Dec 14 '18 at 21:21

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