Chicken Delight claims that 90 percent of its orders are delivered within 10 minutes of the time the order is Data: n 100 p 0.9 p' 0.82 (1) Formulate the hypotheses: H0: p Ha: p < 0.9 0.9 (2) Decide the test statistic and the level of significance: z (Left-tailed), = 0.1 Critical z- score = -1.2816 (3) State the decision Rule: Reject H0 if z < -1.2816 (4) Calculate the value of test statistic: SE = {(p (1 - p)/n} = 0.0300 z = (p' - p)/SE = -2.6667 (5) Compare with the critical value and make a decision: Since -2.6667 < -1.2816 We reject H0 and accept Ha Decision: It appears that less than 90% of the orders are delivered in 10 minutes or less. (6) Calculate the p- value for the test and interpret it: p- value = 0.0038 This is the probability of rejecting a true null hypothesis. 10 minutes of the time the order is placed. A sample of 100 orders revealed that 82 were delivered within the promised timefram We reject H0 and accept Ha d in 10 minutes or less. ecting a true null hypothesis. the promised timeframe. At the 0.10 significance level, can we conclude that less than 90 percent of the orders are delivered in 1 orders are delivered in 10 minutes or less