Statistical test for representativeness of population I have been asked this question:
Suppose a population is 55% female and 45% male. If a sample of 100 yields 53% females and 47% males, can we conclude that the sample is representative of the population.
My way of doing this would be to look at the probability of getting for 53% female versus 55% by chance from this binomial distribution, as follows,
variance = n.p(1-p) = 100 * 0.55 * 0.45 = 24.75
therefore std dev = 4.975
z-score = 53-55 / 4.98 = -0.401
therefore p = 0.655
So because p > 0.05 we cannot reject the null hypothesis that the sample mean = population mean.
Two questions:
1. Is that a valid approach and calculation?
2. In the sample answer I was given they use a chi-squared test, which I've not seen/used before. They give chi-squared value as 0.1616, versus critical value of 3.84 - could someone explain?
Thanks!
 A: With a sample of only $n = 100$ it is difficult to say whether a sample proportion of 53% women is significantly different from a population proportion of 55%. 
To test whether the population from which 100 observations were sampled has fewer the hypothetical population, you would test $H_0: p = .55$ against $H_a: p < .55.$
For $X = 53$ women out of $n = 100,$ an exact binomial test would compute the P-value as $P(X \le 53)$ for $X \sim \mathsf{Binom}(n=100,p=.55).$ In R statistical software, we get P-value 38% which far exceeds 5%.
 pbinom(53, 100, .55)
 [1] 0.3804352

To reject $H_0,$ you would need $X \le 47$ women in your sample of 100.
qbinom(.05, 100, .55)
[1] 47

In the figure below the sum of the heights of the bars to the left of the vertical red line is the P-value.

Note: You might want to check your use of normal tables and investigate using a continuity correction. A more accurate continuity-corrected normal approximation of the P-value than you have computed is 38.2%, which is not much different from the exact binomial value.
pnorm(53.5, 55, sqrt(55*.45))
[1] 0.3815123
pnorm(-1.5/4.975)
[1] 0.3815138

