UPS currently believes that it achieves a 96 % on-time performance for its customers. A random sample of 200 customer shipment studies has shown that 188 of them have met the on-time delivery promise.What is the probability that a random sample of 200 customer shipments would contain 188 or fewer on-time deliveries if the true population of on-time deliveries is 96%?Would you be surprised if the random sample of 200 customer shipments had 197 on-times deliveries? Support your answer with a probability calculation.Suppose the random sample of 200 customer shipments revealed that 178 were on time. Would such a finding cause you to question UPS’s claim that its on-time performance is 96%? Support your answer with a probability calculation.

What is the probability that a random sample of 200 customer shipments would contain 188 or fewer on-time deliveries if the true population of on-time deliveries is 96%? Answer: Here the population proportion = P0 = 0.96 Sample size = n = 200 Shipments would contain 188 or fewer on-time deliveries means a sample proportion of 188200 = 0.94 or less P ( p ≤ 0.94 ) =P p - P0P0×(1-P0)n ≤ 0.94 - 0.960.96×(1-0.96)200 =PZ ≤ -0.020.138=P Z ≤ -0.144= 1-0.5557= 0.443 Would you be surprised if the random sample of 200 customer shipments had 197 on-times deliveries? Support your answer with a probability calculation. Here the population proportion = P0 = 0.96 Sample size = n = 200 Shipments would contain 188 or fewer on-time deliveries means a sample proportion of 197200 = 0.985 or less P ( p = 0.98 ) =P p - P0P0×(1-P0)n = 0.985 - 0.960.96×(1-0.96)200 =PZ = 0.0250.138=P Z = 0.138= 0.5517 No, we will not be surprised as their claimed efficiency level (96%) is below the observed value (98.5%)Suppose the random sample of 200 customer shipments revealed that 178 were on time. Would such a finding cause you to question UPS’s claim that its on-time performance is 96%? Support your answer with a probability calculation. We will question UPS if their on-time performance observed from sample of 200 is less than their claimed 96% on-time performance Here we will test the Hypothesis The following information is provided: The sample size is N = 200N=200, the number of favorable cases is X = 178X=178, and the sample proportion is \bar p = \frac{X}{N} = \frac{ 178}{ 200} = 0.89p¯= NX= 178200=0.89, and the significance level is \alpha = 0.05α=0.05 Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: p = 0.96Ho:p=0.96 Ha: p < 0.96Ha:p<0.96 This corresponds to a left-tailed test, for which a z-test for one population proportion needs to be used. Calculating Test Statistics: Based on the information provided, the significance level is \alpha = 0.05α=0.05, and the critical value for a left-tailed test is z_c = -1.64Zc =−1.64. Since it is observed that z = -5.052 < z_c = -1.64z=−5.052<Zc=−1.64, it is then concluded that the null hypothesis is rejected. Using the P-value approach: The p-value is p = 0p=0, and since p = 0 < 0.05p=0<0.05, it is concluded that the null hypothesis is rejected. Conclusion It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population proportion pp is less than p_0p0, at the \alpha = 0.05α=0.05 significance level. Yes, such a finding will cause us to question UPS’s claim that its on-time performance is 96%

Answered: UPS currently believes that it achieves…

College Algebra

1st Edition

ISBN:9781938168383

Author:Jay Abramson

Publisher:Jay Abramson

Chapter9: Sequences, Probability And Counting Theory

Section9.7: Probability

Problem 1SE: What term is used to express the likelihood of an event occurring? Are there restrictions on its...

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UPS currently believes that it achieves a 96 % on-time performance for its customers. A random sample of 200 customer shipment studies has shown that 188 of them have met the on-time delivery promise.

What is the probability that a random sample of 200 customer shipments would contain 188 or fewer on-time deliveries if the true population of on-time deliveries is 96%?
Would you be surprised if the random sample of 200 customer shipments had 197 on-times deliveries? Support your answer with a probability calculation.
Suppose the random sample of 200 customer shipments revealed that 178 were on time. Would such a finding cause you to question UPS’s claim that its on-time performance is 96%? Support your answer with a probability calculation.

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