(a). The population is still all days that Parsnip has been alive, and the parameter of...

(a). The population is still all days that Parsnip has been alive, and the parameter of interest is = the proportion of all days in which Parsnip eats all the food he is given. It is conjectured that the proportion of all days that he has been alive that Parsnip eats all the food he is given is 0.80, and of interest is to test this claim versus the alternative that the proportion of all days in which Parsnip eats all the food he is given is less than 0.80. State the appropriate null and alternative hypotheses that should be tested.

(b). A simple random sample of 60 days is selected; 45 of those days Parsnip ate all the food he was given, and the other 15 days he did not eat all the food he was given. If appropriate, use this information to test the hypotheses stated in part (a) at significance level = .10.

Solutions

Expert Solution

Given that,
possibile chances (x)=45
sample size(n)=60
success rate ( p )= x/n = 0.75
success probability,( po )=0.8
failure probability,( qo) = 0.2
null, Ho:p=0.8
alternate, H1: p<0.8
level of significance, α = 0.1
from standard normal table,left tailed z α/2 =1.282
since our test is left-tailed
reject Ho, if zo < -1.282
we use test statistic z proportion = p-po/sqrt(poqo/n)
zo=0.75-0.8/(sqrt(0.16)/60)
zo =-0.968
| zo | =0.968
critical value
the value of |z α| at los 0.1% is 1.282
we got |zo| =0.968 & | z α | =1.282
make decision
hence value of |zo | < | z α | and here we do not reject Ho
p-value: left tail - Ha : ( p < -0.96825 ) = 0.16646
hence value of p0.1 < 0.16646,here we do not reject Ho
ANSWERS
---------------
a.
null, Ho:p=0.8
alternate, H1: p<0.8
b.
test statistic: -0.968
critical value: -1.282
decision: do not reject Ho
p-value: 0.16646
we do not have enough evidence to support the claim that the proportion of all days in which Parsnip eats all the food he is given is less than 0.80.

orchestra answered 1 year ago

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