Question

A sous chef in a fancy restaurant boasts of cooking steak perfectly 96% of the time. However one evening, he has already undercooked 3 out of the 20 orders and the orders have been returned to the kitchen. How likely is this? Assuming an a of 0.01, is there sufficient evidence from this evening to suggest that his claim is true?

Answer 1

Let X denote the number of steak that is not properly cooked. Then $X\sim Bin(n=20,p=0.04)$

So, $P(X=x)=\binom{20}{x}0.04^{x}(1-0.04)^{20-x},x=0,1,2,3,.....,20$

Required probability = $P(X=3)=\binom{20}{3}0.04^{3}(1-0.04)^{20-3}=0.0364$

Let p and P denote the sample and population proportion of steak that is not properly cooked respectively. Then,

H0: $P=0.04$

Ha: $P\neq 0.04$

Now, under H0, $p\sim N(P,\frac{P(1-P)}{n})\sim N(0.04,\frac{0.04(1-0.04)}{20})$

Test statistic value = $z=\frac{p-P}{\sqrt{\frac{P(1-P)}{n}}}=\frac{\frac{3}{20}-0.04}{\sqrt{\frac{0.04(1-0.04)}{20}}}=2.51$

Since this is two-tailed test, so p-value is given by: $2\times P(Z>2.51)=2\times [1-P(Z<2.51)]=2\times [1-0.9940]=0.012$

Since p-value = 0.012 is greater than 0.01, so we do not have sufficient evidence to reject H0. Thus we say, $P=0.04$ i.e we say his claim is true.