I'm struggling to figure out how to approach this problem: A tailor thinks that about $75\%$ of customers are satisfied with service. A random sample of $15$ customers, $8$ said they were satisfied. Calculate whether this is plausible at $5\%$ significance level.
I figured that $\mu = 0.6$, and $\bar x = 8/15$, but I'm not sure how to get the variance. Can you assume satisfaction follows a Bernoulli distribution and calculate variance as $(0.75(1-0.75))/n$?
With such a small sample, a normal approximation is inappropriate. Instead, the exact binomial probability is simultaneously easier to understand and guarantees the correct Type I error control.
Consider a rephrasing of the question as follows. You sample $n = 15$ customers, and you observed $X = 8$ satisfied customers. Under the null hypothesis that the true probability of customer satisfaction is $p = 0.75$, what is the probability of obtaining a result at least as extreme as the one observed, $X = 8$?
To compute this, you first compute $$\Pr[X = 8 \mid p = 0.75] = \binom{15}{8} (0.75)^8 (0.25)^7 \approx 0.0393205.$$ Now, among all the other possible outcomes $X \in \{0, 1, 2, \ldots, 15\}$, which of these will have lower probability? Since $8/15 < 0.75$, we know that $X \in \{0, 1, 2, \ldots, 7\}$ will each have smaller probability than $\Pr[X = 8]$. But what about the upper tail? The best way is to compute from $X = 15$ down. We have
$$\Pr[X = 15 \mid p = 0.75] \approx 0.0133635 \\ \Pr[X = 14 \mid p = 0.75] \approx 0.0668173$$
and we stop here because this probability already exceeds $\Pr[X = 8 \mid p = 0.75]$. So the exact $p$-value is $$\begin{align} p &= \Pr[X \le 8 \mid p = 0.75] + \Pr[X = 15 \mid p = 0.75] \\ &\approx 0.0566203 + 0.0133635 \\ &= 0.0699838. \end{align}$$ This exceeds the significance level $\alpha = 0.05$, so the observed data does not furnish enough evidence to reject the hypothesis that $p = 0.75$ at the $5\%$ significance level.
Note that various scientific calculators can compute binomial CDFs, or you can compute them manually; e.g. $$\Pr[X \le 8 \mid p = 0.75] = \sum_{x=0}^8 \binom{15}{x} (0.75)^x (0.25)^{15-x},$$ or you can use a computer, e.g. Wolfram Alpha.