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In a survey conducted by a mail order company a random sample of $200$ customers yielded $172$ who indicated that they were highly satisfied with the delivery time of their orders. Calculate an approximate $95$% confidence interval for the proportion of the company’s customers who are highly satisfied with delivery times.

So I did an approximation using the Central Limit Theorem assuming each observation to be Bernoulli, and I guess assuming the mean and variance of the sum was unknown.

So I used $$\bar{x} \pm 1.96 * \sqrt{{s\over{n}}}$$

The answers however say to use Binomial($200$, $p$), estimating $p$ and finding a confidence interval for that and using $Var(p) = np(1-p)$ so I guess they are assuming the variance is "known". I.e. they don't use sample variance, rather calculate $\sigma^2$ from the binomial distribution after estimating $p=\hat{x}$.

From what I understand my method doesn't assume we know the variance, but does assume a distribution (Bernoulli). Assuming Binomial seems a bigger stretch though. Am I wrong?

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The definition of the Bernoulli Distribution is (as defined by Wikipedia):

In probability theory and statistics, the Bernoulli distribution, (named after Swiss scientist Jacob Bernoulli), is the probability distribution of a random variable which takes the value 1 with probability $p$ and the value 0 with probability $q=1-p$ i.e., the probability distribution of any single experiment that asks a yes–no question; the question results in a boolean-valued outcome, a single bit of information whose value is success/yes/true/one with probability p and failure/no/false/zero with probability q

The definition of the Binomial Distribution (ibid.)

In probability theory and statistics, the binomial distribution with parameters $n$ and $p$ is the discrete probability distribution of the number of successes in a sequence of n independent experiments, each asking a yes–no question, and each with its own boolean-valued outcome

Being that your question deals with a proportion of $179$ "successes" out of $n=200$ "trials". It would seem appropriate to use the Central Limit theorem with parameters based on the Binomial Distribution, since the Bernoulli only deals with individual trials.

So to answer the question we can estimate parameters

$$\hat{p}={179\over{200}}$$ $$\bar{x}=n \hat{p}=179$$ $$s^2=n \hat{p}(1-\hat{p})=18.795 \Rightarrow s \approx 4.335$$

Now to determine the $95$% Confidence Interval

$$\bar{x} \pm 1.96* \left({4.335 \over \sqrt{200}} \right)$$

Which is approximately $$\left [ 178.399, 179.601\right ]$$

Hope this is helpful!

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