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3 years ago

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A recent report from the American Medical Association claims that for the first time in 10 years, the average salary of psychiat

rists was $189,121 with a standard deviation of $26,975. A random sample of 64 psychiatrists this year yielded an average salary of $198,630. Is there evidence that the average salary is actually higher than what the American Medical Association reported

1 answer:

bekas [8.4K]3 years ago

5 0

Answer:

Test statistic

<em> Z = 2.820 > 1.96 at 0.05 level of significance</em>

<em>Null hypothesis is rejected at 0.05 level of significance</em>

<em>The average salary is actually higher than the reported the average of American Medical Association</em>

<u>Step-by-step explanation:</u>

<u><em>Step(i)</em></u>:-

Given data

sample size 'n' = 64

mean of the Population( μ) = $189,121

Standard deviation of the Population(σ) = $26,975

Mean of the sample (x⁻) = $198,630

<u><em>Step(ii):</em></u>-

<u><em>Null hypothesis : H₀ </em></u><em>: there is no significance difference between the means</em>

<u><em>Alternative Hypothesis :H₁</em></u> : <em>The average salary is actually higher than the reported the average of American Medical Association</em>

<em>Test statistic </em>

<em> </em> $Z = \frac{x^{-} -mean}{\frac{S.D}{\sqrt{n} } }$ <em></em>

<em> </em> $Z = \frac{198630 -189121}{\frac{26,975}{\sqrt{64} } }$ <em></em>

<em> Z = 2.820</em>

<em>Level of significance </em>

<em> ∝ =0.05</em>

$Z_{\alpha } = Z_{0.05} =1.96$ <em> </em>

<u><em>Final answer</em></u><em>:-</em>

<em> Z = 2.820 > 1.96 at 0.05 level of significance</em>

<em>Null hypothesis is rejected at 0.05 level of significance</em>

<em>The average salary is actually higher than the reported the average of American Medical Association</em>

<em> </em>

<em></em>

<em></em>

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