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) is that the proportion of customers who arrive late, $p \neq 0.1$.

To find the critical regions at a 5% significance level for a two-tailed test, we determine the threshold for rejection in each tail. Given the null hypothesis proportion $p_0 = 0.1$ and sample size $n = 50$, we find:

1. Calculate the standard error (SE): $SE = \sqrt{\frac{p_0(1 - p_0)}{n}} = \sqrt{\frac{0.1 \times 0.9}{50}} = 0.0424$

2. Determine the critical z-values for a 5% level:

   • Lower critical value: $z_{0.025} \approx -1.96$
   • Upper critical value: $z_{0.975} \approx 1.96$

3. Calculate the critical limits:

   • Lower limit: $np_0 - z_{0.025} \times SE \approx 50 \times 0.1 - 1.96 \times 0.0424 \approx 3.16$
   • Upper limit: $np_0 + z_{0.975} \times SE \approx 50 \times 0.1 + 1.96 \times 0.0424 \approx 6.84$

Thus, we reject the null hypothesis if the number of late customers is less than 4 or more than 6.

The actual level of significance corresponds to the probability of observing a sample statistic in the critical region. Since we found the critical region requires fewer than 4 or more than 6 late customers, we can calculate:

1. Probability for fewer than 4 late arrivals using a binomial distribution: $P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$ Each of these probabilities follows: $P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$ For $p = 0.1, n = 50$:
   • $P(X = 0) \approx 0.0059$
   • $P(X = 1) \approx 0.0323$
   • $P(X = 2) \approx 0.0861$
   • $P(X = 3) \approx 0.1445$

Summing these gives an approximate actual significance level of $0.0003 + 0.0059 + 0.0323 + 0.0861 \approx 0.1243$.

Given that 15 out of the 50 customers arrived late, which is outside the critical region determined in part (b), we see that the observed result supports the manager's belief that the proportion of late arrivals may have changed. In fact, it indicates a higher proportion of late arrivals compared to the previous record of 10%. Therefore, there is evidence against the null hypothesis that the proportion of late arrivals is still 10%.