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?

a

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  

   The sample size is  

    The sample mean is  

    The standard deviation

    The level of significance is  

Now the

    Null  Hypothesis is  

    Alternative Hypothesis is  

generally the degree of freedom is mathematically represented as

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

substituting values

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

687.09 dollars       happy now

step-by-step explanation: 6870

1857 times .63 equals 1169.91 subtract that from 1857 you get 687.09