Question

In an article appearing in Today’s Health a writer states that the average number of calories in a serving of popcorn is 75. To determine if the average number of calories in a serving of popcorn is different from 75, a nutritionist selected a random sample of 20 servings of popcorn and computed the sample mean number of calories per serving to be 78 with a sample standard deviation of 7.
Required:
a. Compute the z or t value of the sample test statistic.
b. At the α = 0.05 level of significance, does the nutritionist have enough evidence to reject the writer's claim?

Answer 1

Answer:

a

$t = 1.9166$

b

The nutritionist does not have sufficient evidence to reject the writers claim

Step-by-step explanation:

From the question we are told that

   The population mean is  $\mu = 75$

   The sample size is  $n = 20$

    The sample mean is  $\= x = 78$

    The standard deviation $\sigma = 7$

    The level of significance is   $\alpha = 0.05$

Now the

    Null  Hypothesis is  $H_o : \mu = 75$

    Alternative Hypothesis is  $\mu \ne 75$

generally the degree of freedom is mathematically represented as

     $df = n-1$

    $df = 20 -1$

    $df = 19$

For a significance level of 0.05 and 19 degrees of freedom, the critical value for the t-test is 2.093. This is obtain from the t-distribution table

The test statistic is mathematically evaluated as

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

substituting values

        $t = \frac{78 - 75 }{\frac{7 }{\sqrt{20} } }$

       $t = 1.9166$

Since the test statistic is below the critical value then the Null hypothesis can not be rejected

So we can conclude that the nutritionist does not have sufficient evidence to reject the writers claim