From 1995 - 2008 in the US 648 people were struck by lightening. 531 of them were men. You want to see if this observation diffe
rs from the null hypothesis, that 50% of people who are hit by lightening would be men, based on the proportion of men in the population as a whole. Calculate the probability under the normal distribution of getting a result of 531 or more (including the continuity correction). Use the binomial test with a normal approximation. What is your conclusion?
1 answer:
Answer:
Because z is higher than any given value in the chart we come to the conclusion To reject null hypothesis
Step-by-step explanation:
Sample proportion = p= 531/648 = 0.8194
This is the proportion of men that were hit by lightening
Null hypothesis: H0: p = 0.5
Alternate hypothesis: H1: p ≠ 0.5
Test statistics z = 0.8194-0.5/(√0.5x0.5/648)
= 0.8194-0.5/√0.0003858
= 0.3194/0.019642
= 16.26
Since the z > 1.96 (at 5% significance) we reject the null hypothesis.
Therefore in conclusion we say z is higher than given values in the chart so we reject null hypothesis.
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For example P(Y=0) =
Like that P(y=1) = 0.1563, P(Y=2) = 0.2891, P(y=3) = 0.2615, P(Y=4) = 0.1516,
P(Y=5) = 0.0667, P(Y=6) = 0.0391,P(Y=7) = 0.0067, P(Y=8)=0.004 and P(Y=9) =0.
a) P(Y>5) = P(Y=6)+P(Y=7)+P(Y=8)+P(Y=9)
= 0.0391+0.0067+0.004+0 = 0.0498
b) P(2<Y<6) = 0.2615+0.1516+0.0667 = 0.4798
c) P(Y≥3) = 0.2615+0.1516+0.0667+0.0391+0.0067+0.004+0= 0.5296
d)P(Y<6) = 0.02493+0.1563+0.2891+0.2615+0.1516+0.0667 =0.95013
e)If we add (a) and (d), we will get 0.0498+0.95013 = 0.99993≈1
Not surprised,since this is nothing but addition of probabilities for all Y values.
That's we got 1 since numerator and denominator are same.