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This can be solved by using the normal approximation to the binomial distribution.
mean = np = 10.000 * 0.5 = 5,000
The standard deviation is given by:
The probability of obtaining more than 5100 tails is 0.0228 and the probability of obtaining fewer than 5100 tails is 0.9772.
The odds of obtaining more than 5100 tails is therefore:
0.0228:0.9772 = 1:42.86.
mean = np = 10.000 * 0.5 = 5,000
The standard deviation is given by:
The probability of obtaining more than 5100 tails is 0.0228 and the probability of obtaining fewer than 5100 tails is 0.9772.
The odds of obtaining more than 5100 tails is therefore:
0.0228:0.9772 = 1:42.86.
Answer:
The P-value is 0.0166.
Step-by-step explanation:
<u>The complete question is:</u> In a one-tail hypothesis test where you reject H0 only in the lower tail, what is the p-value if ZSTAT = -2.13.
We are given that the z-statistics value is -2.13 and we have to find the p-value.
Now, the p-value of the test statistics is given by the following condition;
P-value = P(Z < -2.13) = 1 - P(Z 2.13)
= 1 - 0.9834 = <u>0.0166</u>
Assuming that the level of significance is 0.10 or 10%.
The decision rule for rejecting the null hypothesis based on p-value is given by;
- If the P-value of the test statistics is less than the level of significance, then we have sufficient evidence to reject the null hypothesis.
- If the P-value of the test statistics is more than the level of significance, then we have insufficient evidence to reject the null hypothesis.
Here, the P-value is more than the level of significance as 0.0166 > 0.10, so we have insufficient evidence to reject the null hypothesis, so we fail to reject the null hypothesis.