## Question 265:

1## Answer:

No answer provided yet.### Files Available For Download

So this is a 1 tailed test since we're only interested in whether there is sufficient evidence that patients are paying more than $250 out of pocket. 25 cases have been sampled, which is a decent amount. We want to see if the mean of 275 is sufficient evidence given that we've only sampled 25 cases. We're getting close to the convergence of t-scores and z-scores I'll present just the t-score answer here.- The null hypothesis is that the mean out of pocket expense is less than or equal to $250: Ho: μ ≤ $250.
- The alternative hypothesis is that the mean out of pocket expense is greater than $250 Ha: μ > $250.
- The test statistic t* is derived by subtracting (275.66-250) =25.66 and dividing by the standard error of the mean (SE).
- The SE is the standard deviation divided by the square root of the sample size = 78.11/SQRT(25) = 78.11/5 = 15.622.
- The test statistic t* is 25.66/15.622 = 1.643.
- To find the p-value we use the excel function =TDIST(1.643, 24, 1) or we use the Inverse t-Distribution Calculator and we get .0567, which is just a bit higher than the .05 α cut-off.
- Since the p-value of .0567 is above .05 we do not reject the NULL hypothesis and cannot conclude that the average expense is above $250.
- This is a close decision since the p-value was just barely above .05 (in real life this would likely be sufficient evidence).