Chicken Delight claims that 91 percent of its orders are delivered within 10 minutes of the time the order is placed. A sample of 70 orders revealed that 60 were delivered within the promised time. At the .02 significance level, can we conclude that less than 91 percent of the orders are delivered in less than 10 minutes?(a) What is the decision rule? (Negative amount should be indicated by a minus sign. Round your answer to 2 decimal places.)
Reject Ho if z <
(b)
Compute the value of the test statistic. (Negative amount should be indicated by a minus sign. Round your answer to 2 decimal places.)
Value of the test statistic
(a) To determine the decision rule, we need to find the critical value for a one-tailed test at a significance level of 0.02.
Since the claim is that 91% of the orders are delivered within 10 minutes, the null hypothesis (Ho) is that p = 0.91 (where p represents the proportion of orders delivered within 10 minutes).
The alternative hypothesis (Ha) is that p < 0.91, as we want to conclude that less than 91% of the orders are delivered within 10 minutes.
To find the critical value, we can use a z-table or a statistical calculator. At a significance level of 0.02 (or 2%), the critical value is -2.05 (rounded to 2 decimal places). Therefore, the decision rule is to reject Ho if the test statistic (z) is less than -2.05.
(b) To compute the value of the test statistic, we can use the following formula:
z = (p̂ - p) / √(p * (1 - p) / n)
Where:
- p̂ is the sample proportion (60/70 = 0.857)
- p is the claimed proportion (0.91)
- n is the sample size (70)
Plugging in the values:
z = (0.857 - 0.91) / √(0.91 * (1 - 0.91) / 70)
Now, calculate the numerator:
0.857 - 0.91 = -0.053
Calculate the denominator:
√(0.91 * (1 - 0.91) / 70) = 0.044 (rounded to 3 decimal places)
Now substitute the values:
z = -0.053 / 0.044
Calculate:
z ≈ -1.20 (rounded to 2 decimal places)
Therefore, the value of the test statistic is approximately -1.20.