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Sagot :
Solution :
Given :
[tex]$\mu = 6883$[/tex]
[tex]$s=2518$[/tex]
n = 40
[tex]$\overline x = 5980$[/tex]
Therefore, determining the null hypothesis and alternate hypothesis is :
[tex]$H_0: \mu = 6883$[/tex]
[tex]$H_a: \mu < 6883$[/tex]
The value of the test statistics is :
[tex]$t=\frac{\overline x - \mu}{s/\sqrt n}$[/tex]
[tex]$t=\frac{5980 - 6883}{2518/\sqrt 40}$[/tex]
[tex]$t \approx -2.27$[/tex]
So, df = n - 1
= 40 - 1
= 39
∴ 0.01 < P < 0.025
If the P value is less than α = 0.03, we reject the null hypothesis. Therefore, we conclude that mean annual Medicare spending in Indianapolis is lower than the national mean.
Using the value of [tex]$\alpha = 0.03$[/tex], the critical value of for the test statistics is 2.17
There we reject the null hypothesis. And thus we conclude that the mean annual Medicare spending in Indianapolis is lower than the national mean.
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