1-PropZTeat experienced headaches (based on data from the manufacturer) The acco
ID: 3313986 • Letter: 1
Question
1-PropZTeat experienced headaches (based on data from the manufacturer) The accompanying calculator display shows results from a test of the claim that less than 10% of treated subjects experienced headaches. Use the normal distribution as an approximation to the binomial distribution and assume a 0.05 significance level to complete parts (a) through (e) below. z=-0.447891449 p=0.3271 157700 P 0.0921501706 n- 293 a. Is the test two-tailed, left-tailed, or right-tailed? O Right tailed test O Left-tailed test O Two-tailed test b. What is the test statistic? Z= (Round to two decimal places as needed.) c. What is the P-value? P-value = Round to four decimal places as needed) d. What is the null hypothesis, and what do you conclude about it? Identify the null hypothesis plan O A. Ho:p>0. OB, Ho: pExplanation / Answer
Given that,
possibile chances (x)=27
sample size(n)=293
success rate ( p )= x/n = 0.0922
success probability,( po )=0.1
failure probability,( qo) = 0.9
null, Ho:p=0.1
alternate, H1: p<0.1
level of significance, = 0.05
from standard normal table,left tailed z /2 =1.64
since our test is left-tailed
reject Ho, if zo < -1.64
we use test statistic z proportion = p-po/sqrt(poqo/n)
zo=0.09215-0.1/(sqrt(0.09)/293)
zo =-0.4479
| zo | =0.4479
critical value
the value of |z | at los 0.05% is 1.64
we got |zo| =0.448 & | z | =1.64
make decision
hence value of |zo | < | z | and here we do not reject Ho
p-value: left tail - Ha : ( p < -0.44789 ) = 0.32712
hence value of p0.05 < 0.32712,here we do not reject Ho
ANSWERS
---------------
a.left tailed
b. test statistic: -0.45
c.p-value: 0.3271
d. Ho:p=0.1
e. Option C
f. Option A
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