Statistics Note Sheet
Practice Question #1: A gambler claims he can predict the roll of a die more often than chance would predict. To test this, a die is rolled 100 times and the gambler guesses the number that comes up twenty times and guesses incorrectly the other eighty times. Is this strong evidence that the gambler’s claim is true?
a. Specify the null and alternative hypothesis for this problem. p = probability gambler guess correctly for an individual roll null (Ho): p = 1/6 alternative (Ha): p > 1/6
b. Find the test statistic and calculate the p-value. What do you conclude? pˆ = 0.2 so z = (0.2-0.1667)/ .1667(.8333) /100 = 0.894. From the table the p-value is between 18% and 19%. The null hypothesis is a reasonable explanation of this data so we do not have strong evidence that the gambler can predict the outcome of the die. c. If the gambler had guessed correctly on forty of the rolls, would the p-value go up or down? Since this would be farther away from what is wexpected under the null, the p-value would get smaller. Practice Question #2: Does the person paying the bill order more expensive meals, or less expensive meals, at a restaurant? The bills from 100 parties of two are examined and it turns out that the person paying the bill ordered a meal costing an average of $0.50 more than their companion with a standard deviation of $2.00
a. Clearly define the parameter of interest in this situation and then state the null and alternative hypotheses as statements about this parameter. μ = the average amount the bill-payer orders more than their companion for the whole population…null (Ho): μ = 0 alternative (Ha): μ ≠ 0
b. Find the test statistic and calculate the p-value. What do you conclude? x = 0.5 and z = (0.5-0)/(2.00/100 ) =2.5. From the tables the p-value ≈ 2(100-99.38) = 1.24%. It is unlikely that average differences of this size would occur for 100 pairs of customers. We have evidence that the person paying the bill and their companion do not order the
a. Specify the null and alternative hypothesis for this problem. p = probability gambler guess correctly for an individual roll null (Ho): p = 1/6 alternative (Ha): p > 1/6
b. Find the test statistic and calculate the p-value. What do you conclude? pˆ = 0.2 so z = (0.2-0.1667)/ .1667(.8333) /100 = 0.894. From the table the p-value is between 18% and 19%. The null hypothesis is a reasonable explanation of this data so we do not have strong evidence that the gambler can predict the outcome of the die. c. If the gambler had guessed correctly on forty of the rolls, would the p-value go up or down? Since this would be farther away from what is wexpected under the null, the p-value would get smaller. Practice Question #2: Does the person paying the bill order more expensive meals, or less expensive meals, at a restaurant? The bills from 100 parties of two are examined and it turns out that the person paying the bill ordered a meal costing an average of $0.50 more than their companion with a standard deviation of $2.00
a. Clearly define the parameter of interest in this situation and then state the null and alternative hypotheses as statements about this parameter. μ = the average amount the bill-payer orders more than their companion for the whole population…null (Ho): μ = 0 alternative (Ha): μ ≠ 0
b. Find the test statistic and calculate the p-value. What do you conclude? x = 0.5 and z = (0.5-0)/(2.00/100 ) =2.5. From the tables the p-value ≈ 2(100-99.38) = 1.24%. It is unlikely that average differences of this size would occur for 100 pairs of customers. We have evidence that the person paying the bill and their companion do not order the