Statistical significance?

Question

I have a company's satisfaction score for two different years. Past studies have shown that when a satisfaction score has increased over the past year and is also higher than the national average ($75.7$), then there is a good chance that the stock performs well.

The scores for Year 1 and Year 2 are $73 \& 76$ respectively. Assume that these were derived from a poll with $60$ participants, and historical data shows that the standard deviation is $6$.

Is the increase in score from Year 1 to Year 2 statistically significant with $\alpha = 0.05$?

Now, what exactly is the hypothesis to be tested here? I imagine that the null hypothesis is $$H_0: \text{the increase in score had no positive effect on the stock's performance}$$ and the alternative hypothesis is $$H_1: \text{the increase in score had a positive effect on the stock's performance}$$ But given the data supplied I have no data regarding the stock's performance? Am I misreading the question?

Answer 1

Assuming that data are data are nearly normal and that 'participant' means company, and because the population standard deviation $\sigma = 6$ is given, you can test your null hypothesis $H_0: \mu_2 = \mu_2$ against $H_1: \mu_2 > \mu_1,$ using a two-sample z test.

The test statistic is $$Z = \frac{\bar X_1 - \bar X_1}{\sigma\sqrt{\frac{1}{n_1} + \frac {1}{n_2}}} = \frac{76 - 73}{6\sqrt{2/60}} = \frac{3}{1.095445} = 2.738613 > 1.645.$$

I will leave it to you to verify from your text or class notes that this is the correct test statistic for a 2-sample z test, explain why the denominator is the standard deviation of the numerator (so that the $Z$-statistic is normally distributed if $H_0$ is true), and to explain how the critical value 1.645 arises.

Answer 2

Yes you are misreading the question. The question is: is 76 significantly higher than 73 (so reflecting an actual increase in satisfaction among the bigger population of customers(?) from which the 60 are randomly selected) or can the difference in 3 points be perfectly explained by the random variation that you inevitably get when only asking 60 random participants instead of everybody?

The null hypothesis is thus that in the larger population from which your 60 participants are randomly drawn nothing changes between the two years. This is the same as BruceET's $\mu_1 = \mu_2$: $\mu_i$ is the 'true' satisfaction score (i.e. the average satisfaction score over the entire population that you try to estimate from sample a small subset) in year $i$.

The relation to stock is just there as motivation but does not bear on the actual question.

PS: perhaps you noted that I keep stressing that the 60 is a random sample from a larger population but that is because that is (in my opinion) the most important thing to understand about statistics: you try to get information about a large group of people by looking at a small subset of them. The reason probability theory plays a role is because we model the selection of the small subset from the large population as a random process (drawing balls from an urn) and this is the only source of randomness: the satisfaction $\mu$ in the population is just an existing well defined number, the only problem is that we don't know it.