A dentist knows from past records that 10% of customers arrive late for their appointment - Edexcel - A-Level Maths Statistics - Question 4 - 2022 - Paper 1

The alternative hypothesis (H₁) suggests that there has been a change in the proportion of customers who arrive late. This can be represented as:

$H_1: p \neq 0.1$

To find the critical region at a 5% level of significance, we consider both tails of the distribution. The critical z-values for a two-tailed test are approximately ±1.96. Therefore, if we denote the sample mean proportion as ( ar{p} ) and our sample size as ( n = 50 ), we can use:

$z = \frac{\bar{p} - 0.1}{\sqrt{\frac{0.1(1-0.1)}{50}}}$

The critical region corresponds to rejecting the null hypothesis if (|z| > 1.96). This implies:

• Lower Critical Region: ( p < 0.052 )
• Upper Critical Region: ( p > 0.148 ) (since 0.1 ± 0.048)

Calculating the actual level of significance involves determining the probability of observing 15 or more late customers in the sample of 50, which translates to:

1. Sample proportion: ( ar{p} = \frac{15}{50} = 0.3 )
2. We compare this with our critical regions to find the p-value. The z-score for ( ar{p} ) can be calculated as:

$z = \frac{0.3 - 0.1}{\sqrt{\frac{0.1(1-0.1)}{50}}} \approx 4.47$

Calculating the corresponding p-value (two-tailed) yields an actual significance level of approximately 0.00002, which is significant enough to reject the null hypothesis.

Given that the calculated p-value is well below the significance level of 0.05 and falls within the critical region found in part (b), we have sufficient evidence to support the manager's belief that there has been a change in the proportion of customers arriving late for their appointments.