Question

When testing gas pumps for​ accuracy, fuel-quality enforcement specialists tested pumps and found that 1345 of them were not pumping accurately​ (within 3.3 oz when 5 gal is​ pumped), and 5637 pumps were accurate. Use a 0.01 significance level to test the claim of an industry representative that less than​ 20% of the pumps are inaccurate. Use the​ P-value method and use the normal distribution as an approximation to the binomial distribution.

z=

t=

please expalin steps to take and also if you know the steps on ti-84 calculator please

Answer 1

Z TEST FOR PROPORTION

Given that,
possibile chances (x)=1345
sample size(n)= (1345 + 5637) = 6982
success rate ( p )= x/n = 0.1926
success probability,( po )=0.2
failure probability,( qo) = 0.8
null, Ho:p=0.2  
alternate, an industry representative that less than​ 20% of the pumps are inaccurate, H1: p<0.2
level of significance, alpha = 0.01
from standard normal table,left tailed z alpha/2 =2.326
since our test is left-tailed
reject Ho, if zo < -2.326
we use test statistic z proportion = p-po/sqrt(poqo/n)
zo=0.19264-0.2/(sqrt(0.16)/6982)
zo =-1.5378
| zo | =1.5378
critical value
the value of |z alpha| at los 0.01% is 2.326
we got |zo| =1.538 & | z alpha | =2.326
make decision
hence value of |zo | < | z alpha | and here we do not reject Ho
p-value: left tail - Ha : ( p < -1.53785 ) = 0.06204
hence value of p0.01 < 0.06204,here we do not reject Ho
ANSWERS
---------------
null, Ho:p=0.2
alternate, H1: p<0.2
test statistic: -1.5378
critical value: -2.326
decision: do not reject Ho
p-value: 0.06204
no evidence that an industry representative that less than​ 20% of the pumps are inaccurate