Confidence interval for standard deviation of small sample test when the unknown population var are equal

Question

I have a question about attaining the confidence interval of the standard deviation

Question:

Periodically, customers of a brokerage firm are asked to evaluate their financial consultants. Higher ratings on client satisfaction surveys indicate better perceived service, with 7 the maximum service rating. Independent samples of service rating for two financial consultants are summarized here. Consultant A has 10 years of experience, while consultant B has 1 year of experience $$ \begin{align} Consultant A |& Consultant B \\ \hline n_1 = 16 |& n_2 = 10\\ \bar{X}_1 = 6.82 |& \bar{X}_2 = 6.25 \\ s_1 = 0.64 |& s_2 = 0.75 \end{align} $$

My implementation

$$ \S_p \pm \frac{\bar{X} - \bar{Y}}{ \sqrt{(1/n+1/m)}t_{24,.05/2}} $$

Do I use the pool estimator as the standard deviation?

Answer 1

Unclear what you are trying to do. (1) Test whether mean ratings for A and B are significantly different, which I would guess from the problem you state. Or (2) assume equal variances and get a CI for $\sigma^2 = \sigma_A^2 = \sigma_B^2.$

(1) I see no reason to assume equal variances. To test $H_0: \mu_A = \mu_B$ against $H_a: \mu_A \ne \mu_B$ (or maybe $\mu_A > \mu_B$), I would use a Welch, separate variances t test (output below from Minitab). The CI matching a two-sided test is also given in the printout below. The P-value 6.4% > 5% indicates no significant difference. Accordingly, the 95% CI includes 0, again indicating no significant difference.

Two-Sample T-Test and CI 

Sample   N   Mean  StDev  SE Mean
1       16  6.820  0.640     0.16
2       10  6.250  0.750     0.24

Difference = μ (1) - μ (2)
Estimate for difference:  0.570
95% CI for difference:  (-0.036, 1.176)
T-Test of difference = 0 (vs ≠): T-Value = 1.99  P-Value = 0.064  DF = 16

Note: You can easily tell that this is a Welch t test because a pooled t test would have $DF = 16 + 10 - 2 = 24,$ whereas the printout has $DF = 16.$

(2) Assuming equal population variances (both equal to $\sigma^2$), compute the pooled variance $S_p^2$ and use $24S_p^2/\sigma^2 \sim Chisq(24)$ to get a 95% CI for $\sigma^2.$ I doubt this is good statistical practice, but if you were told to assume equal variances, and then to base a CI for $\sigma^2$ on that assumption, then this would be a way to comply.