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Sagot :
Solution:
To test the hypothesis is that the mean ozone level is different from 4.40 parts per million at 1% of significance level.
The null hypothesis and the alternative hypothesis is :
[tex]$H_0: \mu = 4.40$[/tex]
[tex]$H_a: \mu \neq 4.40$[/tex]
The z-test statistics is :
[tex]$z=\frac{\overline x - \mu}{\left( \sigma / \sqrt n \right)} $[/tex]
[tex]$z=\frac{4.2 - 4.4}{\left(0.7 / \sqrt{23} \right)} $[/tex]
[tex]$z =\frac{-0.2}{0.145}$[/tex]
z = -1.37
The z critical value for the two tailed test at 99% confidence level is from the standard normal table, he z critical value for a two tailed at 99% confidence is 2.57
So the z critical value for a two tailed test at 99% confidence is ± 2.57
Conclusion :
The z values corresponding to the sample statistics falls in the critical region, so the null hypothesis is to be rejected at 1% level of significance. There is a sufficient evidence to indicate that the mean ozone level is different from 4.4 parts per million. The result is statistically significant.
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