Question

To evaluate the effect of a treatment, a sample of n = 9 is obtained from a population with a mean of μ = 40, and the treatment is administered to the individuals in the sample. After treatment, the sample mean is found to be M = 33. If the sample has a standard deviation of s = 9, do we reject or accept the null hypothesis using a two-tailed test with alpha = .05?

Answer 1

Answer:

Yes we reject the null hypothesis

Step-by-step explanation:

From the question we are told that

  The sample size is  $n = 9$

   The  population mean is  $\mu = 40$

    The  sample  mean is  $\= x = 33$

    The  standard deviation is $\sigma = 9$

    The  level of significance is  $\alpha = 0.05$

For a two-tailed test

The  null hypothesis is  $H_o : \mu = 40$

The alternative hypothesis is  $H_a : \mu \ne 40$

Generally the test statistics is mathematically represented as

        $t = \frac{\= x - \mu }{ \frac{\sigma}{\sqrt{n} } }$

=>     $t = \frac{ 33 - 40 }{ \frac{9}{\sqrt{9} } }$

=>     $t = -2.33$

The  p-value for the two-tailed test  is mathematically represented as

     $p-value = 2 P(z > |-2.33|)$

From the z-table  

            $P(z > |-2.33|) = 0.01$

       $p-value = 2 * 0.01$

       $p-value = 0.02$

Given that $p-value < \alpha$ Then we reject the null hypothesis