To confirm if the clinic is indeed meeting its goal of maintaining average satisfaction levels above 7, we perform a hypothesis test. Let's walk through the steps.
The null hypothesis is that the average satisfaction is 7, while the alternative hypothesis states that the satisfaction is more than 7.
1. Firstly, we have to compute a z-score. This score is a measure of how many standard deviations an element is from the mean. We do this by subtracting the null hypothesis mean (7, in this case) from the sample mean (7.8) and divide the result by the standard deviation (2.3) divided by the square root of the sample size (23). Our calculated Z score yields 1.6681153124565977.
2. Secondly, we calculate a p-value. This is the probability that a random chosen patient has a score greater than what the Z-Score states if the null hypothesis is assumed to be true. Ideally, we want to achieve as small a p-value as possible to confidently reject the null hypothesis.
To derive the p-value, we look at the right tail of the normal distribution since the Z-Score is positive and our alternative hypothesis states that the satisfaction is more than the stated null hypothesis (7 in this case). The p-value is 1 minus the cumulative distribution function (the integral from -infinity to the Z-Score under the normal distribution curve) of the Z-Score. This calculation gives us a p-value equal to 0.04764641901172839.
3. The p-value is less than the usual significance level of 0.05. The significance level is the probability of rejecting the null hypothesis if it is true. For a significance level of 0.05, if the null was true, we would expect to get this sample data or something more extreme 5% of the time.
From these calculations, we conclude to reject the null hypothesis because our p-value of 0.04764641901172839, which is less than the significance level 0.05. This indicates that we have sufficient evidence to conclude that the average satisfaction level of patients at the clinic is indeed above 7.