Complete Question
The complete question is shown on the first uploaded image
Answer:
a
Yes the researcher can conclude that the supplement has a significant effect on cognitive skill
b
[tex]d = 0.5778 [/tex]
c
The result of this hypothesis test shows that there is sufficient evidence to that the supplement had significant effect.The measure of effect size is large due to the large value of Cohen's d (0.5778 > 0.30 )
Step-by-step explanation:
From the question we are told that
The sample size is n=16
The sample mean is [tex]M = 50.2[/tex]
The standard deviation is [tex]\sigma = 9[/tex]
The population mean is [tex]\mu = 45[/tex]
The level of significance is [tex]\alpha = 0.05[/tex]
The null hypothesis is [tex]H_o \mu =45[/tex]
The alternative hypothesis is [tex]H_a \mu \ne 45[/tex]
Generally the test statistics is mathematically represented as
[tex]z = \frac{M - \mu}{\frac{\sigma}{\sqrt{n} } }[/tex]
=> [tex]z = \frac{50.2 - 45}{\frac{9}{\sqrt{16} } }[/tex]
=> [tex]z = 2.31 [/tex]
Generally the p-value is mathematically represented as
[tex]p-value = 2 * P(Z > z )[/tex]
[tex]p-value = 2 * P(Z > 2.31 )[/tex]
From the z-table
[tex]P(Z > 2.31 ) = 0.010444[/tex]
=> [tex]p-value = 2 * 0.010444 [/tex]
=> [tex]p-value = 0.021 [/tex]
From the obtained values we see that [tex]p-value < 0.05[/tex]
Decision Rule
Reject the null hypothesis
Conclusion
There is sufficient evidence to conclude that the supplement has a significant effect on the cognitive skill of elderly adults
Generally the Cohen's d for this study is mathematically represented as
[tex]d = \frac{M - \mu}{\sigma }[/tex]
=> [tex]d = \frac{50.2 -45}{9 }[/tex]
=> [tex]d = 0.5778 [/tex]
