STAT - Statistics Project Example

From these results we reject the null hypothesis that the average hotel price in Philadelphia is $129.50 since we have proven that the average hotel room at Philadelphia is significantly different from $125 hence its not $ 129.50. Sample size = 250 and they believed that the average is 75%. It means that the confidence interval is at α = 0.25. We calculate the sample mean as xbar= 250/168 = 1.488 and that the claimed average =0.75 thus t = 1.48-0.75)/ Since the resultant Z statistic test is less than the obtained value of (168/250) = 0.

672, we reject the null hypothesis that there proportion of drivers under influence of alcohol are significantly less than 75%. We conclude that the proportion of drivers under the influence of alcohol is not less than 75%. Z = {Sample mean- μ0}/ δ/√n = {9-6} ÷ 2.47/√100 = 12.146. The critical value of Z is obtained from the tables as Z α /2 = Z 0.05 /2 = Z 0.025 = 1.96. but since the assumed is greater than the test statistic, we reject the null hypothesis that the mean waiting time will be 6 minutes or that its not true that the average waiting time will be reduced from the current 9- 10 minute.

Let us find out the critical value of chi-squared (χ2) = (n-1) s2 ÷ δ2 = (25-1)0.4 /0.3 = 32. This is the obtained chi-square and from the chi-square tables, we deduce the value of the critical region as 36.42 (Bowman par 3). Since it’s a left- tailed test, our rejection region is less than 36.42 i.e. (χ2