Final answer:
To determine whether there is sufficient evidence to support the bride's hope that the average cost of a wedding is less than the reported average, we can conduct a hypothesis test. The null hypothesis is that the average cost of a wedding is equal to the reported average ($26,716), and the alternative hypothesis is that it is less than the reported average. Using a one-sample t-test, we compare the sample mean ($25,637) to the population mean ($26,716) to calculate the t-value. Comparing the t-value to the critical t-value at the 0.05 level of significance, we can reject the null hypothesis and support the bride's hope.
Step-by-step explanation:
To determine whether there is sufficient evidence to support the bride's hope that the average cost of a wedding is less than the reported average, we can conduct a hypothesis test. The null hypothesis is that the average cost of a wedding is equal to the reported average ($26,716), and the alternative hypothesis is that it is less than the reported average. We can compare the sample mean ($25,637) to the population mean ($26,716) using a one-sample t-test.
Using the formula for the t-statistic: t = (sample mean - population mean) / (sample standard deviation / sqrt(sample size)), we can calculate the t-value. In this case, the t-value is approximately -1.977.
Using a t-table or a statistical software, we can find the critical t-value for a one-tailed test at the 0.05 level of significance and 56 degrees of freedom (sample size - 1). The critical t-value is approximately -1.673. Since the t-value (-1.977) is less than the critical t-value (-1.673), we can conclude that there is sufficient evidence to support the bride's hope. Therefore, the answer is A) Yes, reject the null hypothesis.