Answer:
The value of the test statistic to 2 d.p is z= 1.65
Step-by-step explanation:
P cap= 0.72
n= 170
P= 0.66
q= 1- p
q= 1- 0.66
q= 0.34
Z=( p cap - p)/√(p*q)/n
Z= (0.72- 0.66)/√(0.66*0.34)/170
Z= 0.06/0.036332
Z= 1.65
See attached picture
Answer:
Null hypothesis: H0 = 0.72
Alternative hypothesis: Ha <> 0.72
z score = −1.74
P value = P(Z<-1.74) + P(Z>1.74) = 0.041 + 0.041= 0.082
Step-by-step explanation:
Given;
n=170 represent the random sample taken
Null hypothesis: H0 = 0.72
Alternative hypothesis: Ha <> 0.72
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 = 0.72
p^ = Observed proportion = 0.66
Substituting the values we have
z = (0.66-0.72)/√{0.72(1-0.72)/170}
z = −1.74233013109
z = -1.74
To determine the p value (test statistic) at 0.05 significance level, using a two tailed hypothesis. Using the p- value table;
P value = P(Z<-1.74) + P(Z>1.74) = 0.041 + 0.041= 0.082
Since z at 0.05 significance level is between -1.96 and +1.96, and the z score for the test (z = -1.74) 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.
