The answer is that the Governor's claim is correct.
We know that ,
X - μ / s/√n
t ₙ₋₁ i.e; t-distribution with (n-1) d.f. when the population standard deviation is unknown.
X - t₀.₉₅,ₙ₋₁ / s/√n
X + t₀.₉₅,ₙ₋₁ / s/√n
Let n be the sample size of the given sample.
The 95% confidence interval of the population mean is where,
n = 78, s = 3.
Given the sample size, we can compute the above.
be the average income of nurses.
We have to test, H0:
= 55,775 against H1:
> 55,775
The test-statistic is, T = X - μ / s/√n
X = 58,445,
X = 55,775, s = 3800, n = 60
Thus, T = 5.4426
Under H0, T ~ t-distribution with (n-1) d.f. i.e. 59 d.f.
The p-value = P(T > 5.4426) = 0
Since, p-value < level of significance (0.05), we reject the null hypothesis H0.
So, we conclude that the Governor's claim is correct.
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