Guns in the Home in a Gallup poll of 1003 randomly selected subjects, 373 said t

Question

Guns in the Home in a Gallup poll of 1003 randomly selected subjects, 373 said that they have a gun in their home. Test of the claim that 35% of homes have guns in them Use alpha = 0.05 and provide the hypotheses alpha level, rejection region, calculation, p-value. result, and conclusion Also compute a 95% confidence interval about the population proportion and state whether this confidence interval is equivalent to the test of hypothesis and explain your answer. If using the Binomial distribution, what would be the exact p-value? H_0: H_1: alpha = Rejection Region/Critical Region: Reject H_0 if Otherwise, fail to reject. Calculations/Test Statistic: p-value = Result: Reject H_0 at the alpha = level of significance. Fail to reject H_0 at the alpha = level of significance. Conclusion 95% Cl: Binomial p-value

Explanation / Answer

PART A.
Given that,
possibile chances (x)=373
sample size(n)=1003
success rate ( p )= x/n = 0.3719
success probability,( po )=0.35
failure probability,( qo) = 0.65
null, Ho:p=0.35
alternate, H1: p!=0.35
level of significance, = 0.05
from standard normal table, two tailed z /2 =1.96
since our test is two-tailed
reject Ho, if zo < -1.96 OR if zo > 1.96
we use test statistic z proportion = p-po/sqrt(poqo/n)
zo=0.37188-0.35/(sqrt(0.2275)/1003)
zo =1.4531
| zo | =1.4531
critical value
the value of |z | at los 0.05% is 1.96
we got |zo| =1.453 & | z | =1.96
make decision
hence value of |zo | < | z | and here we do not reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 1.45309 ) = 0.1462
hence value of p0.05 < 0.1462,here we do not reject Ho
ANSWERS
---------------
null, Ho:p=0.35
alternate, H1: p!=0.35
test statistic: 1.4531
critical value: -1.96 , 1.96
decision: do not reject Ho
p-value: 0.1462

PART B.
Confidence Interval For Proportion
CI = p ± Z a/2 Sqrt(p*(1-p)/n)))
x = Mean
n = Sample Size
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
CI = Confidence Interval
No. of success(x)=373
Sample Size(n)=1003
Sample proportion = x/n =0.372
Confidence Interval = [ 0.372 ±Z a/2 ( Sqrt ( 0.372*0.628) /1003)]
= [ 0.372 - 1.96* Sqrt(0) , 0.372 + 1.96* Sqrt(0) ]
= [ 0.342,0.402]
Interpretations:
1) We are 95% sure that the interval [0.342 , 0.402 ] contains the true population proportion
2) If a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the true population proportion  

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