Answer:
[tex]z=\frac{0.21 -0.15}{\sqrt{\frac{0.15(1-0.15)}{100}}}=1.68[/tex]
Step-by-step explanation:
Information provided
n=100 represent the random sample taken
X=21 represent the number of bags overfilled
[tex]\hat p=\frac{21}{100}=0.21[/tex] estimated proportion of overfilled bags
[tex]p_o=0.15[/tex] is the value that we want to test
z would represent the statistic
Hypothesis
We need to conduct a hypothesis in order to test if the true proportion of overfilled bags is higher than 0.15.:
Null hypothesis:[tex]p =0.7[/tex]
Alternative hypothesis:[tex]p > 0.15[/tex]
The statistic for this case is:
[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)
And replacing the info given we got:
[tex]z=\frac{0.21 -0.15}{\sqrt{\frac{0.15(1-0.15)}{100}}}=1.68[/tex]