If I know that the actual proportion of faulty appliances is 0.05 for the population, and that for some sample of 1000 there is 100 faulty appliances. Then what would the approximate probability be that there is less than 100 faulty appliances in an arbitrary sample of 1000?
I've thought of assuming a binomial distribution:
$P(X \le 100) = \sum_{k=0}^{100}$ $1000 \choose k $ $(0.1)^{k} (0.9)^{1000-k}$
$\approx 0.99999999996091791276975545969378411463693255629937624686$
Is this the right way of going about the problem?
Thanks in advance.
It seems there is doubt that the sample of 1000, comes from the population with 5% faulty appliances. Let the null hypothesis be $H_0: p = .05.$
If $H_0$ is true then the number $X$ of faulty items in a sample of $n = 1000$ has $X \sim \mathsf{Binom}(n = 1000, p = .05).$ So we would 'expect' about $E(X) = np = 50$ defective items in a random sample of size $n = 1000,$ whereas we find $100.$
You should compute $P(X \ge 100) = 1 - P(x \le 99) = 1 - \sum_{k=0}^{99}$ $1000 \choose k $ $(0.05)^{k} (0.95)^{1000-k} \approx 0.$ Using R statistical software, I get:
So if the sample were really taken from the hypothetical population, there is almost no chance of seeing 100 or more defectives. This is the 'P-value' of a test of $H_0$ against the alternative $H_a: p > .05.$ It is the probability of a result as extreme or more extreme--in the direction of the alternative--than what was observed.
We reject $H_0$ at any reasonable level of significance. Data (100 defectives) are not consistent with $H_0.$ If people are claiming the sample came from a population with $p = .05,$ it would not be wise to believe them.
Here is a plot of the PDF of the null distribution $\mathsf{Binom}(1000, .05).$ The P-value is the probability to the right of the dotted vertical line.