Statistics: What is the probability of type I error?

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SUMMARY

The discussion focuses on calculating the probabilities of type I and type II errors for a binomial distribution with parameters n=10 and p=0.6. The rejection region for the null hypothesis H_0: p=0.6 is defined as {X ≤ 1} ∪ {X ≥ 9}. The calculated probability of type I error is approximately 0.1689, while the probability of type II error, when H_1 is changed to H_1: p=0.3, is approximately 0.617. A participant reported different values of 0.0480 for type I error and 0.8585 for type II error, indicating a potential discrepancy in calculations.

PREREQUISITES
  • Understanding of binomial distribution and its parameters (n and p).
  • Knowledge of hypothesis testing concepts, specifically type I and type II errors.
  • Familiarity with binomial probability formulas, including the use of combinations.
  • Ability to perform calculations involving summation of probabilities for discrete distributions.
NEXT STEPS
  • Study the derivation and application of the binomial distribution formula: f(x) = \binom{n}{x} p^x(1-p)^{n-x}.
  • Learn about hypothesis testing frameworks, focusing on type I and type II error calculations.
  • Explore statistical software tools like R or Python for performing binomial probability calculations.
  • Investigate the impact of sample size and parameter changes on error probabilities in hypothesis testing.
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Students studying statistics, data analysts, and researchers involved in hypothesis testing and statistical inference.

sanctifier
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Homework Statement



X is a random variable of binomial distribution of parameter n=10 and unknown parameter p. Hypotheses are given as follows:

H_0 \;\; : \;\; p=0.6

H_1 \;\; : \;\; p \neq 0.6

Suppose rejection region for H_0 is \{X \leq 1\} \cup \{X \geq 9\}

Question 1: What is the probability of type I error?

Question 2: If H_1 is changed to "H_1 \;\; : \;\; p =0.3", then what is the probability of type II error?

Homework Equations



Binomial Distribution of parameters n and p: f(x) = \binom{n}x p^x(1-p)^{n-x}

The Attempt at a Solution



Answer 1: Desired probability is

P(X \leq 1,\;\; X \geq 9|p=0.6)=1-P(2\leq X \leq 8|p=0.6)= 1-\sum_{k=2}^8 \binom{10}k 0.6^k(1-0.6)^{10-k} \approx 0.1689

Answer 2:

P(2 \leq X \leq 8|p=0.3) = \sum_{k=2}^8 \binom{10}k 0.3^k(1-0.3)^{10-k} \approx 0.617

Are these answers correct? Thank you in advance!
 
Last edited:
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Help!

Does anyone know the correct solution?
 
sanctifier said:
Help!

Does anyone know the correct solution?

I get 0.04803512320 ≈ 0.0480 for question 1 and .8505479682 ≈ 0.8585 for question 2.
 

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