in 22 minutes with a standard deviation of 5 minutes. You want to test their claim at a significance level of 0.05.
Here's how we will proceed:
1. We start with the null hypothesis, which is the assumption that the restaurant's claim is true, i.e., the average delivery time is 20 minutes.
2. Then we formulate an alternative hypothesis, which is contrary to the null hypothesis. In this case, the alternative hypothesis will be that the average delivery time is not 20 minutes.
3. We calculate the z-score, which gives us an idea of how far our sample mean (22 minutes) is from the population mean (20 minutes), measured in number of standard deviations. The formula for the z-score is (sample mean - population mean) divided by the standard deviation divided by the square root of the sample size. Accordingly, our z score becomes (22-20) / (5/(36^0.5)), which equals 2.4. This z-score of 2.4 means that the sample mean of 22 minutes is 2.4 standard deviations away from the claimed population mean of 20 minutes.
4. Once we have the z score, we can calculate the p-value. P-value is the probability that, given that the null hypothesis is true, you would get the observed data (or data more extreme). Using the properties of the normal distribution, we find that the p-value for a z-score of 2.4 is approximately 0.0082.
5. We compare the p-value to our chosen significance level of 0.05. If p-value is less than the significance level, we reject the null hypothesis and accept the alternative hypothesis. In our case, since the p-value of 0.008 is less than 0.05, we reject the null hypothesis.
Based on this statistical analysis, it would be right to state that the restaurant's claim of an average delivery time of 20 minutes is likely not true. However, please bear in mind that statistical tests do not always reflect the practical significance of a result, and the small difference in average delivery times might not be significant in practical terms.