Why this implication is true

Question

I'm studying this book and on page 364 when he is starting to define confidence intervals, he made this big jump first saying the sample proportion fall within 1.96 standard deviation of the population proportion (it's natural, since the population proportion is in the center of the distribution) to just after saying the other way around, it's the population proportion that falls within 1.96 standard deviation of the sample proportion.

Why this is true? following the first sentence we should have had:

...,Then the interval from

[population proportion - 1.96(standard deviation)] to

[population proportion + 1.96(standard deviation)]

contains the sample proportion.

I need help, for me this isn't clear at all

Answer 1

If the standard deviation does not change (for example it might be known), the two statements appear logically equivalent to saying that the absolute difference between the sample proportion and the population proportion is less than $1.96$ standard deviations $95\%$ of the time. They are not quite the same:

• One says that given the fixed interval of the population proportion $\pm1.96$ standard deviations, $95\%$ of varying sample proportions will fall in that interval being considered

• The other says that of varying intervals of the sample proportion $\pm1.96$ standard deviations, $95\%$ of such intervals will contain the fixed population proportion , but does not tell you about whether a particular such interval being considered may contain the population proportion

Answer 2

Note that the statement has emphasized "does".

Basically, it is saying that while 95% of the time, sample proportions ($\bar x$) fall within the $1.96\sigma$ range of the population proportion ($\mu$), let's assume it is actually true for some given sample. If so then, the interval given in the statement (the confidence interval) will contain actual population parameter.

You can also think of this in terms of conditional probability. This is how confidence interval is obtained:

$$Pr\bigg(\frac{\bar x - \mu}{\sigma} \in (-1.96, 1.96)\bigg) =0.95$$

Now the statement asserts that conditional probability:

$$Pr\bigg(\frac{\bar x - \mu}{\sigma} \in (-1.96, 1.96) \bigg| \bar x \in (\mu - 1.96\sigma, \mu + 1.96\sigma)\bigg) =1$$

I think you can now easily see why this statement is true.