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Answer:
Step-by-step explanation:
Given that:
Population Mean = 7.1
sample size = 24
Sample mean = 7.3
Standard deviation = 1.0
Level of significance = 0.025
The null hypothesis:
[tex]H_o: \mu = 7.1[/tex]
The alternative hypothesis:
[tex]H_a: \mu > 7.1[/tex]
This test is right-tailed.
[tex]degree \ of \ freedom= n - 1 \\ \\ degree \ of \ freedom = 24 - 1 \\ \\ degree \ of \ freedom = 23[/tex]
Rejection region: at ∝ = 0.025 and df of 23, the critical value of the right-tailed test [tex]t_c = 2.069[/tex]
The test statistics can be computed as:
[tex]t = \dfrac{ \hat X - \mu_o}{\dfrac{s}{\sqrt{n}}}[/tex]
[tex]t = \dfrac{ 7.3-7.1}{\dfrac{1}{\sqrt{24}}}[/tex]
[tex]t = \dfrac{0.2}{0.204}[/tex]
t = 0.980
Decision rule:
Since the calculated value of t is lesser than, i.e t = 0.980 < [tex]t_c = 2.069[/tex], then we do not reject the null hypothesis.
Conclusion:
We conclude that there is insufficient evidence to claim that the population mean is greater than 7.1 at 0.025 level of significance.