A random sample of 18 female neurologists and 21 male neurologists at public medical schools found that the average salary for female neurologists was $175,000 with a standard deviation of $15,000, while the average salary for male neurologists was $203,000 with a standard deviation of $22,000. Find a 99% confidence interval for the difference between the average salaries of male and female neurologists. (round your answer to the nearest dollar. ) assume that both populations are normally distributed. Assume the two population variances are unequal. Which fomula should be used to compute the interval. Use subscript 1 for female and 2 for male.

The confidence interval for the difference between the average salaries of male and female neurologists is (19,131.77, 36,868.23). And the formula which should be used to compute the interval is x₁` - x₂`/ Sp √ (1/n₁ + 1/n₂).

The 99% confidence interval for the difference between the average salaries of male and female neurologists is given by:

(x₁` - x₂`) ± t∝/2(υ) * √ (Sp²/n₁ + Sp²/n₂)

= x₁` - x₂`/ Sp √ (1/n₁ + 1/n₂)

where Sp² = s₁² (n₁ - 1) + s₂² (n₂ - 1) / (n₁ - 1) + (n₂ - 1)

The sample should be normally distributed, and it is assumed that the population standard deviations are known, which is sometimes met and sometimes not known.

Now, we have H0 : μ₁ = μ₂ against the claim Ha : μ₁ ≠ μ₂.

And the test statistic is:

t = x₁` - x₂` / √ (Sp²/n₁ + Sp²/n₂)

Sp² = s₁² (n₁ - 1) + s₂² (n₂ - 1) / (n₁ - 1) + (n₂ - 1)

= 15,000² (17) + 22,000² (22) / 37

= 103,378,640

Now, degree of freedom (df):

df = n₁ + n₂ - 2 = 21 + 18 - 2 = 37

For 2-tailed-test, the critical value of t is obtained by:

t ≥ t∝/2 (df) = t ≥ t0.005 (37) = ± 2.7154 = ± 2.72

Last, insert the values to the formula:

(x₁` - x₂`) ± t∝/2(υ) * √ (Sp²/n₁ + Sp²/n₂)

= 28,000 ± (2.72 * 3,265.89)

= (19,131.77, 36,868.23)

Hence, the confidence interval is (19,131.77, 36,868.23).

Learn more about confidence interval at: https://brainly.com/question/24131141

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