How can I tell which basketball shooting format is better? I am experimenting with two shooting formats in basketball practice with our team.  One format is competitive shooting, and the other one is non-competitive shooting (all same shots, except the motivations are different). I have the following statistics:
                           Makes     Takes     Average     95%-Confidence 
Competitive Shooting        98       165       59.39%      51.9% to 66.9%
NOT Competitive Shooting    522      963       54.21%      51.1% to 57.4%

This is the extent of my knowledge and I'm not sure how to determine whether one format is better than another? Or if I'm using the correct thinking process for this?
 A: One way to look at this is to perform a $\chi^2$ test on the table you presented:
dat <- matrix(c(98, 522, 165, 963), 2)
chisq.test(dat)

Which returns:
    Pearson's Chi-squared test with Yates' continuity correction

data:  dat
X-squared = 0.34761, df = 1, p-value = 0.5555

This indicates that there is no significant difference between the two types of shooting. Since you have a theoretical independent variable and dependent variable, however, you could also re-create a full data set and perform a logistic regression. I think this is a little bit more intuitive:
format <- factor(c(rep("competitive_yes", 263), rep("competitive_not", 1485)))
make <- c(rep(TRUE, 98), rep(FALSE, 165), rep(TRUE, 522), rep(FALSE, 963))
mod <- glm(make ~ format, family = binomial)
round(summary(mod)$coef, 3)

Which returns pretty much the same p-value for significance:
                      Estimate Std. Error z value Pr(>|z|)
(Intercept)             -0.612      0.054 -11.267     0.00
formatcompetitive_yes    0.091      0.139   0.659     0.51

Logistic regression makes things linear by forcing things into log-odds units (or logits), so you can use the exponential function to get the odds ratio, which in this case is exp(.091) or $1.10$. So competitive format shots are 1.1 times as likely to be made as non-competitive shots, but this is not statistically significant, $p = .510$. That is to say, if there was no difference between the two conditions, we would expect to get an odds ratio of 1.10 or higher 51% of the time. In fact, if we ran this study an arbitrarily large number of times, we would expect to see the odds ratio be between 0.83 and 1.44 95% of the time (which I calculated using the Std. Error).
