Question

A survey conducted by the American Automobile Association showed that a family of four spends an average of $215.60 per day while on vacation Suppose a sample of 64 families of four vacationing at Niagara Falls resulted in a sample mean of $252.45 per day and a sample standard deviation of $79.50 Devo a 95% confidence interval estimate of the mean amount spent per day by a family of four visiting Niagara Falls (to 2 decimals). . Based on the confidence interval from part (a), does it appear that the population mean amount spent per day by families visiting Niagara Falls differs from the mean reported by the American Automo

Answer 1

Let i denotes the sample mean, s denotes the sample standard deviation and n denotes the sample size. Here, 79.500 Ž = 252.450, s2 = 6320.250000, s = 79.500, n = 64, standard error = - V64 = 9.937500 9.938 alue is tcritical = t 1-0.95,63 = to.025,63 = 1.998341 ~ 1.998 At 95 percent confidence level, The margin of error is: 79.500 ME= 79.500 * 1.998 = 19.875000 ~ 19.875 * t 1-0.95,63 = * 10.025,63 = V64 V64 a 95 percent confidence interval of population mean u is Sets - *t 1-0.95,63 vn 20 = 252.450 79.500 - - * to.025,63 764 79.500 = 252.450 +- * 1.998 V64 = 252.450 + 19.875 = (232.575000, 272.325000) ~ (232.57, 272.32)

Since the confidence interval does not include 215.60, so at 95% level of confidence, we can conclude that population mean amount spent per day by families visiting Niagra falls is significantly different from the mean reported by American automobiles.