Understanding Hypothesis Testing and Power

This problem asks us to calculate the power of a specific hypothesis test related to the probability of getting a head in a coin toss. We are given the null hypothesis ($H_0$) and the alternative hypothesis ($H_1$), the number of coin tosses, and the rule for rejecting $H_0$.

Defining the Hypotheses and Test

• The null hypothesis is $H_0: p = 0.4$. This is the default assumption about the probability of getting a head.
• The alternative hypothesis is $H_1: p = 0.6$. This is the claim we are trying to find evidence for.
• We perform an experiment consisting of 6 coin tosses ($n=6$).
• The test's decision rule is to reject $H_0$ if the number of heads observed in the 6 tosses is five or more (i.e., $\ge 5$).

What is the Power of a Test?

The power of a hypothesis test is the probability of correctly rejecting the null hypothesis ($H_0$) when the alternative hypothesis ($H_1$) is actually true. In simpler terms, it's the probability that our test will detect a true effect (in this case, that the probability of heads is indeed 0.6) when that effect exists.

Mathematically, Power $= P(\text{Reject } H_0 | H_1 \text{ is true})$.

Calculating the Power for this Coin Toss Test

To find the power, we need to calculate the probability of rejecting $H_0$ assuming that $H_1$ is true. According to $H_1$, the true probability of a head is $p = 0.6$.

The number of heads in a fixed number of independent coin tosses follows a binomial distribution. Here, the number of tosses is $n=6$. If $H_1$ is true, the probability of success (getting a head) is $p=0.6$. Let $X$ be the number of heads in 6 tosses. Under $H_1$, $X \sim B(n=6, p=0.6)$.

The rejection region is $X \ge 5$. So, we need to calculate $P(X \ge 5 | p=0.6)$. This is the sum of the probabilities of getting exactly 5 heads or exactly 6 heads in 6 tosses, with $p=0.6$.

The probability mass function for a binomial distribution is given by:

\( P(X=k) = \binom{n}{k} p^k (1-p)^{n-k} \)

Where:

• \(n = 6\) (number of tosses)
• \(p = 0.6\) (probability of head under \(H_1\))
• \(k\) is the number of heads

Step 1: Calculate the probability of getting exactly 5 heads (\(P(X=5)\))

\( P(X=5) = \binom{6}{5} (0.6)^5 (1-0.6)^{6-5} = \binom{6}{5} (0.6)^5 (0.4)^1 \)

\( \binom{6}{5} = 6 \)

\( (0.6)^5 = 0.07776 \)

\( (0.4)^1 = 0.4 \)

\( P(X=5) = 6 \times 0.07776 \times 0.4 = 6 \times 0.031104 = 0.186624 \)

Step 2: Calculate the probability of getting exactly 6 heads (\(P(X=6)\))

\( P(X=6) = \binom{6}{6} (0.6)^6 (1-0.6)^{6-6} = \binom{6}{6} (0.6)^6 (0.4)^0 \)

\( \binom{6}{6} = 1 \)

\( (0.6)^6 = 0.046656 \)

\( (0.4)^0 = 1 \)

\( P(X=6) = 1 \times 0.046656 \times 1 = 0.046656 \)

Step 3: Calculate the total probability of rejecting \(H_0\) under \(H_1\)

Power \( = P(X \ge 5 | p=0.6) = P(X=5) + P(X=6) \)

Power \( = 0.186624 + 0.046656 = 0.23328 \)

Rounding to three decimal places, the power of the test is approximately 0.233.

Conclusion on Test Power

The calculated power of the test is 0.233. This value represents the probability of detecting that the true probability of heads is 0.6 when it is indeed 0.6, given the test setup (6 tosses, reject if $\ge 5$ heads).

Revision Table: Key Concepts for Hypothesis Testing Power

Additional Information on Power and Sample Size

The power of a test is influenced by several factors:

• Sample Size (n): Increasing the sample size generally increases the power of the test, assuming the effect size is real. With more data, it's easier to distinguish between the null and alternative hypotheses. In our coin toss example, increasing the number of tosses would likely increase the power to detect a difference between p=0.4 and p=0.6, provided the rejection region is adjusted appropriately.
• Significance Level (\(\alpha\)): Increasing the significance level (\(\alpha\)) increases the power of the test. This is because a larger \(\alpha\) makes it easier to reject \(H_0\), thus increasing the chance of correctly rejecting \(H_0\) when \(H_1\) is true. However, it also increases the probability of a Type I error.
• Effect Size: The magnitude of the difference between the null hypothesis value and the true value under the alternative hypothesis is the effect size. A larger effect size (e.g., the difference between 0.4 and 0.6 is larger than the difference between 0.4 and 0.45) is easier to detect, leading to higher power.
• Variability: Lower variability in the data increases the power of the test.

Understanding and calculating power is crucial in study design, as it helps determine the sample size needed to have a reasonable chance of detecting a statistically significant result if one exists.