9.22 The mean arrival rate of flights at O’Hare Airport in marginal weather is 195 flights per hour with a historical standard deviation of 13 flights. To increase arrivals, a new air traffic control procedure is implemented. In the next 30 days of marginal weather the mean arrival rate is 200 flights per hour. (a) Set up a right-tailed decision rule at a=.025 to decide whether there has been a significant
increase in the mean number of arrivals per hour. (b) Carry out the test and make the decision. Is it close? Would the decision be different if you used a = .01? (c) What assumptions are you making, if any?
Flights
210 215 200 189 200 213 202 181 197 199
193 209 215 192 179 196 225 199 196 210
199 188 174 176 202 195 195 208 222 221
Solution:
(a) Null Hypothesis: H0: 0= 195 vs. Ha:a195
Rejection region can be defined as
z >z
Here significance level is 0.025,
So, z0.025= 1.96[critical value](Using Statistical Ratio Calculator
From for calculating z with 0.025 significance)
(b)Now,
z = (X - μ) / σx
Where X is a normal random variable, μ is the mean, and σ is the standard deviation.
Where n is the sample size.
Calculating t-test statistics:
z = 200-195/(13/√30) = 2.106625
Calculating p-value:
Degree of freedom = DF = 30-1 = 29
P(z292.106625) = 0.021961(Using t=2.106625and DF = 29)
The p-value of 2.196% is less than significant level of 2.5%and 2.106625 is greater than critical value 1.96. Hence, we can reject the null hypothesis and conclude that there has been a significant increase in the mean number of arrivals per hour.
When significance level is 0.01,
So, z0.01= 2.3263[critical value](Using Statistical Ratio Calculator
From for calculating z with 0.01 significance)
The p-value of 2.196% is less than significant level of 1.0% and 2.106625 is less than critical value 2.3263. Hence, we can accept the null hypothesis and conclude that there has been no increase in the mean number of arrivals per hour.
(c)I am assuming the sample is selected independently and randomly from population. Population size is sufficiently large in sample.
9.54 Faced with rising fax costs, a firm issued a guideline that transmissions of 10 pages or more should be sent by 2-day mail instead. Exceptions are allowed, but they want the average to be 10 or below. The firm examined 35 randomly chosen fax transmissions during the next year, yielding a sample mean of 14.44 with a standard deviation of 4.45 pages. (a) At the .01 level of significance,
is the true mean greater than 10? (b) Use Excel to find the right-tail p-value.
Solution:
(a) Null Hypothesis: H0: 0= 10 vs. Ha:a10
Rejection region can be defined as
z >z
Here significance level is 0.01,
So, z0.025= 2.3263[critical value](Using Statistical Ratio Calculator
From for calculating z with 0.01 significance)
Now,
z = (X - μ) / σx
Where X is a normal random variable, μ is the mean, and σ is the standard deviation.
Where n is the sample size.
Calculating t-test statistics:
z = 14.44-10/(4.45/√35) = 5.903
Calculating p-value:
Degree of freedom = DF = 35-1 = 34
P(z295.903) = 0.000001 (Using t=5.903and DF = 34)
The p-value of 0.0001% is less than significant level of 1.0% and 5.903is greater than critical value 2.3263. Hence, we can reject the null hypothesis and conclude that true mean is greater than 10.
(b) Use Excel to find the right-tail p-value
Applying the function TDIST(5.903,34,1) in excel gives the value of p as 5.75452E-07.
This is infinitesimally small in comparison of 0.01.
9.56 A coin was flipped 60 times and came up heads 38 times. (a) At the .10 level of significance, is the coin biased toward heads? Show your decision rule and calculations. (b) Calculate a p-value and interpret it.
Solution:
(a) At the .10 level of significance, is the coin biased toward heads? Show your decision rule and calculations.
Null Hypothesis: H0: p0= 0.5 vs. Ha:pa0.5
Rejection region can be defined as
z >z
Here significance level is 0.10,
So, z0.10= 1.2816[critical value](Using Statistical Ratio Calculator
From for calculating z with 0.10 significance)
Now,
z = (X - μ) / σx
Where X is a normal random variable, μ is the mean, and σ is the standard deviation.
Where n is the sample size.
Calculating t-test statistics:
z = 0.633-0.5/(0.5/√60) = 2.066
Calculating p-value:
Degree of freedom = DF = 60-1 = 59
P(z292.066) = 0.021614 (Using t=2.066and DF = 59)
The p-value of 2.1614% is less than significant level of 10.0% and 2.066is greater than critical value 1.2816. Hence, we can reject the null hypothesis and conclude that coin is biased.
(b) Calculate a p-value and interpret it.
Applying the function TDIST(2.066,59,1) in excel gives the value of p as 0.102498428
.
This is greater than significance level. Hence we can reject the null hypothesis.
9.62 The Web-based company Oh Baby! Gifts have a goal of processing 95 percent of its orders on the same day they are received. If 485 out of the next 500 orders are processed on the same day, would this prove that they are exceeding their goal, using a = .025? (See story.news.yahoo.com accessed
June 25, 2004.)
Solution:
P485 = 485/500 = 0.97
Null Hypothesis: H0: p0= 0.95 vs. Ha:pa0.95
Rejection region can be defined as
z >z
Here significance level is 0.025,
So, z0.10= 1.96[critical value](Using Statistical Ratio Calculator
From for calculating z with 0.025 significance)
Now,
z = (p485 - p) / √(p*(1-p))/n
Where n is the sample size.
Calculating t-test statistics:
z =0.97-0.95/√(0.95*0.05/500) = 2.0519
Calculating p-value:
Degree of freedom = DF = 500-1 = 499
P(z292.0519) = 0.020350 (Using t=2.0519and DF = 499)
The p-value of 2.0350% is less than significant level of 2.50% and 2.0519is greater than critical value 1.96. Hence, we can reject the null hypothesis and conclude that they are exceeding their goal.