Question

A restaurant chain regularly surveys its customers. On the basis of these surveys, the management of the chain claims that 75% of its customers rate the food as excellent. A consumer testing service wants to examine the claim by asking 460 customers to rate the food.

Seventy per cent rated the food as excellent. Does this support the management’s claim?

Answer 1

$H_0 : p = 0.75 \\ H_1 : p< 0.75$

σ = 95 %

n = 460

$\bar{x}=0.70 \\ z = \frac{0.70-0.75}{\sqrt{ \frac{0.75 \times 0.25}{460} }} = -2.476 \\ p(-2.4766) = 0.0066$

Since this p-value is less than the level of significance (0.05), the decision is to reject the null hypothesis.