Using the same scenario as above, but now you do not know the population variance. You collect a random sample of n = 21 cereal boxes and find the sample mean is 12.8 and the sample standard deviation to be 0.5. Test the null hypothesis H0: µ = 13 against the alternative hypothesis HA: µ < 13 at level of significance α = 0.01. Required:
a. Find the test statistic.
b. Using the rejection region, do you reject or not reject the null hypothesis?
c. Find the p-value.
d. Using the p-value, do you reject or not reject the null hypothesis?
e. In context of your problem, what is your conclusion about the ounces contained in cereal boxes?

-1.833 ; we do not reject the Null.

Pvalue = 0.0409 ; We do not reject the Null

There is no significant evidence to conclude that the number of ounces contained in the cereal boxes is less than 13

Step-by-step explanation:

H0 : μ = 13

HA : μ < 13

Test statistic :

(xbar - μ) ÷ s/sqrt(n)

(12.8 - 13) ÷ 0.5/sqrt(21)

-0.2 / 0.1091089

= - 1.833

The critical value at α = 0.01 ; df = 21 - 1 = 20 = 2.528

-1.833 > - 2. 528 (left tailed test) ; Hence we do not reject the Null.

Test statistic :

Pvalue from Test statistic at DF = 20 ; α = 0.01

Pvalue = 0.0409

Pvalue > α

0.0409 > 0.01 ; Hence, we do not reject H0.

There is no significant evidence to conclude that the number of ounces contained in the cereal boxes is less than 13

because they could drop even lower..

your answer would be 12