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

answer: 80,000

step-by-step explanation:

the population doubles every day.

10,000*2=20,000 day 4

20,000*2=40,000 day 5

40,000*2=80,000 day 6