Hypothesis Testing (Binomial Distribution) (AQA A-Level Mathematics): Revision Notes

Example: It is claimed in an election that a certain party has 40% of the votes. A sample of 10 people were asked what party they vote for, and only 2 said they would vote for that party. Test whether the claim was true.

Hypothesis:

It is suspected that the party's claim of 40% vote share was overstated.

Note: In order to objectively perform the test, we must have a threshold of likelihood below which we reject a claim. This threshold is called the significance level and is usually given as a percentage probability.

We will perform the test at the 5% significance level, meaning any event cumulatively in the bottom 5% of probabilities will lead to rejection of the 40% claim.

Step 1: Define the variable and parameter $p$ that we are testing

$X$= "the no. of people voting for the party"

$p$ = "the probability of a person voting for the party"

Step 2: Write down the null and alternate hypotheses

$H_0: p = 0.4$ $\leftarrow$ We initially believe the claim

$H_1: p < 0.4$ $\leftarrow$ The alternative given in the question

Step 3: Test the observed data at the given significance level

In order to perform this test, we check whether the observation of 1 person out of 10 lies in the bottom cumulative 5%.

To do this, since we are testing for $p < 0.4$, we look at the probability to the left of our observation:

(Where $X \sim B(10, 0.4)$)

5% Significance Level:

Anything in the bottom 5% (i.e., the rejection region) will lead to a rejection of $H_0$.

Step 4: Compare observed probability to significance level, then conclude

0.04636 < 0.05 (Key phrase to use: "Reject $H_0$")

Sufficient evidence to suggest that the party has less than 40% of the votes.

(This is our alternative hypothesis.)

Common Errors:

• Not initially defining $X$ and $p$ in words.
• Writing $H_0: p$ and $H_1: p$ instead of $H_0: p$ = and $H_1: p$ <.
• When performing the test, using $P(X \leq 1)$ instead of $P(X = 1)$.
• Not comparing probability to significance level.

Explanation:

The test we have just performed is called a one-tail test as we are only testing one end of the distribution (i.e., our rejection region was on the left-hand side/left-hand tail). The following is an example of a right-tail test:

Example: A single observation, $x$, is taken from a binomial distribution $B(10, p)$ and a value of 5 is obtained. Use this observation to test $H_0: p = 0.25$ against $H_1: p > 0.25$ using a 5% significance level.

[Note: since this question has no context, it is not possible to define $X, p$, or give an in-context conclusion.]

$H_0: p = 0.25 \\ H_1: p > 0.25$

Our observed data was 5 out of 10.

Do not reject $H_0$. [Note: "Accept $H_0$" is incorrect.]

In an in-context question, our conclusion would read "insufficient evidence to suggest...".

Example: A dice used in playing a board game is suspected of not giving the number 6 often enough. During a particular game, it was rolled 12 times and only one 6 appeared.

Does this represent significant evidence, at the 5% level of significance, that the probability of a 6 on this dice is less than $\frac{1}{6}$?

$X$ = "the no. of sixes rolled"

$p$ = "probability of rolling a $6$"

$H_0: p = \frac{1}{6} \\ H_1: p < \frac{1}{6}$

$X \sim B(12, \frac{1}{6})$

$P(X \leq 1) = 0.3813$

0.3813 > 0.05 $\checkmark$

Do not reject $H_0$ $\checkmark$

Insufficient evidence to suggest that a 6 being thrown has a probability less than $\frac{1}{6}$.

Example: Binomial Distribution with a 5% Significance Level Given:

• Binomial distribution $B(20, 0.3)$
• 5% significance level

Calculation:

$P(X \leq 0) = 0.000798 (0.01$%)

$P(X \leq 1) = 0.00764 (0.08$%)

$P(X \leq 2) = 0.0355 (3.55$%) $\leftarrow$ This is within the 5% significance level

$P(X \leq 3) = 0.107 (10.7$%) $\rightarrow$ This is outside the 5% significance level

Result:

• Critical Values: 0, 1, and 2
• Critical Region: $X \leq 2$

Example: Testing for Change in the Ratio of Fruit to Mint Flavours at the 10% Significance Level Given:

A sweet manufacturer packs sweets with 70% fruit and the rest mint flavoured. They want to test if there has been a change in the ratio of fruit to mint flavours at the 10% significance level. To do this they take a sample of 20 sweets. What are the critical regions?

$X$= number of fruit sweets. Binomial (20, 0.7)

$H₀: p = 0.7$

$H₁: p ≠ 0.7$

10% Significance level (2 tailed – 5% at each tail)

Lower Tail:

• $P(X \leq 10) = 0.0480 (4.8$%) $\rightarrow$ Within the 5% lower tail Critical Region $X \leq 10$, Critical Value $= 10$

• $P(X \leq 11) = 0.113 (11.3$%) $\rightarrow$ Outside the 5% lower tail

Upper Tail:

• $P(X \geq 17) = 0.107 (10.7$%)
• $P(X \geq 18) = 0.035 (3.5$%) $\rightarrow$ Within the 5% upper tail Critical Region $X \geq 18$, Critical Value $=18$

Result:

• Critical Region $X \leq 10$ or $X \geq 18$
• Critical Values: $X = 10$ and $X = 18$

Example A machine makes glass bowls, and it is observed that one in ten of the bowls has hairline cracks in them. The production process is modified, and a sample of 20 bowls is taken. One of the bowls is cracked. Test, at the 10% level of significance, the hypothesis that the proportion of cracked bowls has changed as a result of the change in the production process. State your hypotheses clearly.

$X$ = "the number of cracked bowls"

$p$ = "the probability of a bowl being cracked"

Null Hypothesis:

$H_0: p = 0.1$

Alternative Hypothesis:

$H_1: p \neq 0.1$ We don't know whether $p > 0.1$ or $p < 0.1$

• Significance Level: 5% + 5% = 10% Note: In performing such a test, we go through broadly the same process as in a one-tail test with a few key differences, which will be highlighted.

Calculation:

• Expected number of cracked bowls in sample: 0.1 × 20 = 2

• We observed 1 breakage < 2 (expected number).

• $p$ sample size.

Test:

• Testing for decrease in p (i.e., left tail): $P(X \leq 1) = 0.392 > 0.05 \quad (\text{Refer to t}_{4})$

Do not reject $H_0$ $\quad\checkmark$ Insufficient evidence to suggest that the number of cracked bowls has changed.

Example: A standard blood test is able to diagnose a particular disease with a probability of 0.96. A manufacturer suggests that a cheaper test will have the same probability of success. It conducts a clinical trial on 75 patients. The new test correctly diagnoses 63 of these patients. Test the manufacturer's claim at the 10% level, stating your hypotheses clearly.

Step 1: Define the variables and hypotheses.

$X$ = "the number of correct diagnoses"

p = "the probability of a correct diagnosis"

Null Hypothesis:

$H_0: p = 0.96$

Alternative Hypothesis:

$H_1: p \neq 0.96$

Step 2: Perform the binomial test.

The number of trials $(n) = 75$, and the success probability under $H_0$ is $p = 0.96$.

This gives:

$X \sim B(75, 0.96)$

Expected number of correct diagnoses = 0.96 × 75 = 72 correct diagnoses.

Observed number of correct diagnoses = 63 (which is less than 72).

We are testing for a decrease (5% each tail).

Step 3: Calculate the probability:

Conclusion:

Reject $H_0$

There is sufficient evidence to suggest that the success rate for the new test is different from before.

Q2 (June 2006, Q2)

Question A

• The random variable $R$ has the distribution $B(6, p)$.
• A random observation of $R$ is found to be 6.
• Test at a 5% significance level the null hypothesis $H_0: p = 0.45$ against the alternative hypothesis $H_1: p \neq 0.45$.

Method:

Calculate the expected value:

Since 6 is greater than the expectation, it's a right-tail test.

Calculate

Compare with the significance level:

Conclusion: Reject $H_0$.

Question B:

• The random variable $S$ has the distribution $B(n, p)$.
• Test at the same significance level as in part (i).
• A random observation of $S$ is found to be 1.
• Use tables to find the largest value of n for which $H_0$ is not rejected.

Method:

• Since 1 is less than the expectation of 2.7, it's a left-tail test.
• Test different values of n to find the largest n where $P(X \leq 1)$ is greater than the significance level:
• $n = 5: P(X \leq 1) = 0.256 > 0.025$ (Not rejected).
• $n = 10: P(X \leq 1) = 0.0233 < 0.025$ (Rejected).
• $n = 9: P(X \leq 1) = 0.0385 > 0.025$ (Not rejected).
• Conclusion: The largest value of $n$ is 9.

Q3 (June 2007, Q7) A Television company believes that the proportion of households that can receive Channel C is 0.35.

Question A:

In a random sample of 14 households it is found that 2 can receive Channel C.

Test, at the 2.5% significance level, whether there is evidence that the proportion of households that can receive Channel C is less than 0.35

Steps:

1. Define the hypotheses:

• $H_0: p = 0.35$
• $H_1: p < 0.35$

2. Model the problem using a binomial distribution: $X \sim B(14, 0.35)$.
3. Calculate the probability $P(X \leq 2) = 0.0839$.
4. Compare this probability to the significance level (0.025):

• 0.0839 > 0.025

5. Conclusion: Do not reject $H_0$. There is insufficient evidence to suggest that the proportion of households receiving Channel C is less than 0.35.

Question B:

On another occasion the test is carried out again, with the same hypotheses and significance level as in part (i), but using a new sample, of size $n$.

It is found that no members of the sample can receive Channel C.

Find the largest value of $n$ for which the null hypothesis is not rejected.

Show all relevant working.

Steps:

6. For each possible sample size $n$, calculate $P(X \leq 0)$.
7. Compare these probabilities to the significance level (0.025):

• $n = 5: P(X \leq 0) = 0.1160 \gt 0.025$ (not rejected)
• $n = 10: P(X \leq 0) = 0.0135 \lt 0.025$ (rejected)
• $n = 9: P(X \leq 0) = 0.0207 \lt 0.025$ (rejected)
• $n = 8: P(X \leq 0) = 0.0319 \gt 0.025$ (not rejected)

8. Conclusion: The largest n for which $H_0$ is not rejected is n = 8.