Question

A quality control manager is concerned about variability of the net weight of his company’s individual yogurt cups. To check the consistency, he takes a random sample of sixteen 6-ounce yogurt cups and finds the mean of the sampled weights to be 5.85 ounces and the sample standard deviation to be 0.2 ounce. Test the hypotheses H0: µ ≥ 6 Ha: µ < 6 at the 5% level of significance. Assume the population of yogurt-cup net weights is approximately normally distributed. Based on the results of the test, will the manager be satisfied that the company is not under-filling its cups? State the decision rule, the test statistic, and the manager’s decision.

Answer 1

Answer:

Decision rule : The  p-value  <  $\alpha$ so  the  null hypothesis is rejected

The test statistics is  $t = -2.8$

The  manger will not be manager be satisfied that the company is not under-filling since the company is under-filling its cups

Step-by-step explanation:

From the question we are told that

    The  sample size is  n  =  16

     The  sample  mean is  $\= x = 5.85$

     The  standard deviation  is  $\sigma = 0.2$

      The  null hypothesis is  $H_o : \mu \ge 6$

       The alternative  hypothesis is  $H_a : \mu < 6$

       The  level of significance is  $\alpha = 0.05$

       

Generally the test statistics is mathematically represented as

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

=>    $t = \frac{ 5.86 - 6 }{ \frac{ 0.2}{ \sqrt{ 16} } }$

=>   $t = -2.8$

The  p-value is obtained from the z-table  the value is  

       $p-value = P(Z < -2.8 ) = 0.0025551$

     $p-value = 0.0025551$

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

Hence there is sufficient evidence to support the concern of the quality control manager. and the manger will not be satisfied that since the test proof that the company is under-filling its cups