I True or False 20 pts Write T if the statement is true and

I. True or False (20 pts) Write T if the statement is true and Fif the statement is false. Also, if the statement is false, underline the word or phrase making the statement incorrect and supply the correct word or phrase to make it true 1. If a hypothesis is rejected at the 5% level of significance, it may or may not be rejected at the 10% level. 2. In order to decrease the length of any confidence interval, there must be a decrease in the sample size or one has to use a smaller confidence evel. 3. We would like to estimate the mean GPA of the population of all PGCC undergraduates. We would like to compute a 90 percent confidence interval for this mean and we would like our interval to have margin of error no greater than 0.1. The registrar\'s office says that a reliable guess ofa population standard deviation is 0.5. The smallest number of students we should study to achieve our goal is 144. 4. A random sample of 400 passengers of an airline is polled after their flights. Of the passengers, 300 say they will fly again with the same airline. At 90% confidence interval for the proportion of passengers that will fly again with the same airline is 0.75 +/-0.066. 5·The quality department of an electronics manufacturer randomly selected 100 resistors. The mean resistance of the resistors was 20 1.5 and the standard deviation was .4 The 98% confidence interval for this problem is 201.5+/-125. 6. If the p - value is 0.1 and the level of significance is 0.05 in a hypothesis test for a mean, then we have strong evidence to accept the alternative hypothesis. 7. A certain confidence interval of population proportion p is 0.35s p

Solution

1. FALSE. Replace \"may or may not\" to \"will be\".

2. FALSE. replace \"decrease in sample size\" to \"increase in sample size.\"

3. FALSE. Replace 144 by 68.

Note that      
      
n = z(alpha/2)^2 s^2 / E^2      
      
where      
      
alpha/2 = (1 - confidence level)/2 =    0.05  
      
Using a table/technology,      
      
z(alpha/2) =    1.644853627  
      
Also,      
      
s = sample standard deviation =    0.5  
E = margin of error =    0.1  
      
Thus,      
      
n =    67.63858635  
      
Rounding up,      
      
n =    68

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