Answer :
Answer:
The value of the test statistic to 2 decimal place is 1.91
Step-by-step explanation:
P cap= 0.42
n= 170
P= 0.35
q= 1-p
q= 1-0.35=0.65
Z=( pcap - p)/√(p*q)/n
Z= (0.42 - 0.35)/ √ (0.35*0.65)/170
Z= 0.07/√1.3383 * 10 ^ -3
Z= 1.91
Therefore the test statistic is z=1.91
See attached picture
Answer:
Null hypothesis: H0 = 0.42
Alternative hypothesis: Ha <> 0.42
z score = −1.849 = -1.85
P value = P(Z<-1.849) + P(Z>1.849) = 0.032 + 0.032= 0.064
Step-by-step explanation:
Given;
n=170 represent the random sample taken
Null hypothesis: H0 = 0.42
Alternative hypothesis: Ha <> 0.42
Test statistic z score can be calculated with the formula below;
z = (p^−po)/√{po(1−po)/n}
Where,
z= Test statistics
n = Sample size = 170
po = Null hypothesized value = 0.42
p^ = Observed proportion = 0.35
Substituting the values we have
z = (0.35-0.42)/√{0.42(1-0.42)/170}
z = -1.849
z = -1.85
To determine the p value (test statistic) at 0.05 significance level, using a two tailed hypothesis.
P value = P(Z<-1.849) + P(Z>1.849) = 0.032 + 0.032= 0.064
Since z at 0.05 significance level is between -1.96 and +1.96, and the z score for the test (z = -1.849) falls with the region bounded by Z at 0.05 significance level. Then we can conclude that we don't have enough evidence to FAIL or reject the null hypothesis, and we can say that at 5% significance level the null hypothesis is valid.