I have a probability and statistics question relating to p-values. The question is from a Master's course. The scenario is the following:
A representative of a potential supplier claims that at least 90% of the equipment manufactured and sold by the company has no defects. You have obtained a sample of equipment usage records from a firm that uses these machines. Of 121 machines in the records, 105 machines have no recorded defects. What is the probability of this event if the supplier’s claim is correct (i.e. The p -value). On the basis of this data, do you believe the representative’s claim?
My thoughts so far:
The significance level of the test: α = 0.1 (10%)
The percentage of 121 machines that have no recorded defects is 86.7%.
I am not sure how to arrive at the p-value from this data.
Any help would be appreciated, thank you.
First of all: the significance level is not given here. They are asking you to calculate the p-value and decide considering this probability. If the p-value is very low, you decide in favour of the alternative hypothsesis. Usually a significance level for the test is 5% and a high significance level is 1%.
You have a sample mean of $\frac{105}{121}\approx 86.78\%$ and you have to verify if this value is far from the claimed $90\%$. Techincally, this is a "One Tailed Test"
To do that, you have to use the well known formula and calculate the probablity of the event (using CLT)
$$Z_{\text{stat}}=\frac{\overline{p}-p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}\dot\sim N(0;1)$$
using your data you get
$$Z_{\text{stat}}=-1.18$$
Now using tables (or calculator) you get
$$p_{\text{value}}=\mathbb{P}[Z<-1.18]\approx 11.86\%$$
which is a high p-value, not enough to reject the claimed null hypothesis. In other words, there is no a significant difference between the claimed 90% and the observed 86.8%