We considered the differences between the reading and writing scores of a random sample of 200 students who took the High School and Beyond Survey. The mean and standard deviation of the differences...

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  1. We considered the differences between the reading and writing scores of a random sample of 200 students who took the High School and Beyond Survey. The mean and standard deviation of the differences are ¯xread-write=-0.545and 8.887 points.


Calculate a 95% confidence interval for the average difference between the reading and writing scores of all students. (Use the z-table for this calculation.)______________________
Does the confidence interval provide convincing evidence that there is a real difference in the average scores?
Yes or no

  1. Identify the critical t. An independent random sample is selected from an approximately normal population with unknown standard deviation. Find the degrees of freedom and the critical t value (t*) for the given sample size and confidence level.


(a) n = 7, CL = 90%
df= ____ t*=____
(b) n = 26, CL = 98%


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We considered the differences between the reading and writing scores of a random sample of 200 students who took the High School and Beyond Survey. The mean and standard deviation of the differences are ¯xread-write=-0.545 and 8.887 points. Calculate a 95% confidence interval for the average difference between the reading and writing scores of all students. (Use the z-table for this calculation.)______________________ Does the confidence interval provide convincing evidence that there is a real difference in the average scores? Yes or no Identify the critical t. An independent random sample is selected from an approximately normal population with unknown standard deviation. Find the degrees of freedom and the critical t value (t*) for the given sample size and confidence level. (a) n = 7, CL = 90% df= ____ t*=____ (b) n = 26, CL = 98% df= ____ t*=____ (c) n = 28, CL = 95% df=____ t*=____ (d) n = 11, CL = 99% df= ____ t*=____ Prices of diamonds are determined by what is known as the 4 Cs: cut, clarity, color, and carat weight. The prices of diamonds go up as the carat weight increases, but the increase is not smooth. For example, the difference between the size of a 0.99 carat diamond and a 1 carat diamond is undetectable to the naked human eye, but the price of a 1 carat diamond tends to be much higher than the price of a 0.99 diamond. In this question we use two random samples of diamonds, 0.99 carats and 1 carat, each sample of size 23, and compare the average prices of the diamonds. In order to be able to compare equivalent units, we first divide the price for each diamond by 100 times its weight in carats. That is, for a 0.99 carat diamond, we divide the price by 99. For a 1 carat diamond, we divide the price by 100. The distributions and some sample statistics are shown below. 0.99 carats 1 carat___ Mean $44.51 $56.81 SD $13.13 $16.13 n 23 23 Conduct...



Answered Same DayDec 26, 2021

Answer To: We considered the differences between the reading and writing scores of a random sample of 200...

Robert answered on Dec 26 2021
120 Votes
1. The 95% Confidence interval for the average difference between the reading and writing
scores of all students is
0.95
8.887
0.
545 1.96
200
0.545 1.23
( 1.775,0.685)
CI x z
n

 
   
  
 
The above interval contains zero, hence the null hypothesis could not be rejected. So, the
confidence interval does not provide convincing evidence that there is a real difference in the
average scores. The answer is NO.
2. Since, independent random samples are selected from an approximately normal population
with unknown standard deviation the degrees of freedom will be n-2.
(a) n = 7, CL = 90%
df= _7-2=5___ t*=__1.476(one tail), 2.015(both tails)__
(b) n = 26, CL = 98%
df= _26-2=23___ t*=__2.177(right tail), 2.5(both tail)__
(c) n = 28, CL = 95%
df=__28-2=26__ t*=__1.706(one tail), 2.056(both tail)__
(d) n = 11, CL = 99%
df= _11-2=9___ t*=_2.821 (one tail), 3.250(both tail)___
3. The null hypothesis is
0 1 2: 0H    (There is no significant difference between the two population means)
Against the alternative hypothesis
1 1 2: 0H    (The mean of the first population is different from the second)
The level of significance is 0.01  .
The degrees of freedom is:
   
   
1 2 1 21 1
23 1 23 1
44
df df df n n     
   


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