Question

Researchers have noted a decline in cognitive functioning as people age (Bartus, 1990). However, the results from other research suggest that the antioxidants in foods such as blueberries can reduce and even reverse these age-related declines, at least in laboratory rats (Joseph et al., 1999). Based on these results, one might theorize that the same antioxidants might also benefit elderly humans. Suppose a researcher is interested in testing this theory. The researcher obtains a sample of adults who are older than 65, and gives each participant a daily dose of a blueberry supplement that is very high in antioxidants. After taking the supplement for 6 months, the participants are given a standardized cognitive skills test and produce a mean score of . For the general population of elderly adults, scores on the test average μ and form a normal distribution with σ=9.
(a)-Can the researcher conclude that the supplement has a significant effect on cognitive skill? Use a two-tailed

Answer 1

Complete Question

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Answer:

a

Yes the researcher can  conclude that the supplement has a significant effect on cognitive skill

b

$d = 0.5778$

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  $M = 50.2$

    The standard deviation is $\sigma = 9$

    The  population mean is  $\mu = 45$

     The level of significance is  $\alpha = 0.05$

The null hypothesis is $H_o \mu =45$

The alternative hypothesis is $H_a \mu \ne 45$

Generally the test statistics is mathematically represented as

          $z = \frac{M - \mu}{\frac{\sigma}{\sqrt{n} } }$

=>       $z = \frac{50.2 - 45}{\frac{9}{\sqrt{16} } }$

=>       $z = 2.31$

Generally the p-value is mathematically represented as

     $p-value = 2 * P(Z > z )$

      $p-value = 2 * P(Z > 2.31 )$

From the z-table  

      $P(Z > 2.31 ) = 0.010444$

=>     $p-value = 2 * 0.010444$

=>     $p-value = 0.021$

From the obtained values we see that $p-value < 0.05$

 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

    $d = \frac{M - \mu}{\sigma }$

=>  $d = \frac{50.2 -45}{9 }$

=>  $d = 0.5778$