Question

The ability to read rapidly and simultaneously maintain a high level of comprehension is often a determining factor in the academic success of many high school students. A school district is considering a supplemental reading program for incoming freshmen. Prior to implementing the program, the school runs a pilot program on a random sample of n 5 20 students. The students were thoroughly tested to determine reading speed and reading comprehension. Based on a fixed-length standardized test reading passage, the following reading times (in minutes) and comprehension scores (based on a 100-point scale) were recorded.

Student 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 n y s

Reading Time 5 7 15 12 8 7 10 11 9 13 10 6 11 8 10 8 7 6 11 8 20 9.10 2.573

Comprehension 60 76 76 90 81 75 95 98 88 73 90 66 91 83 100 85 76 69 91 78 20 82.05 10.88

a. What is the population about which inferences are being made?

b. Place a 95% confidence interval on the mean reading time for all incoming freshmen in the district.

c. Plot the reading time using a normal probability plot or boxplot. Do the data appear to be a random sample from a population having a normal distribution?

d. Provide an interpretation of the interval estimate in part (b).

Step 1

Hello there! there are more than 3 subparts in the question. According to our principles can answer only 3 subparts. Do find the solution to parts a,b,c below. If you have doubts in remaining subparts request as a new quesiton.

Given data of 20 students reading time in minutes and comprehension scores. This data is taken to test the supplemental reading program for incoming freshmen. So here the population is the total high school students in the district.

Step 2

Need to determine 95% confidence interval on the mean reading time for all incoming freshmen in the district. The confidence interval can be found using the below mentioned formula. Here we know sample mean of reading time = 9.1 and sample standard deviaiton = 2.573. n = 20.

given 95% confidence interval so α = 1-0.95 = 0.05.

So (1-α/2) = 1-(0.05/2) = 1-0.025 = 0.975.

degree of freedom = n-1 = 20-1 = 19.

So the critical t score is 2.093 for 19 degree of freedom and 0.025 area to the right.

Step 3

Using the given data we calculate margin of error amd then we caclualte the confidence interva...

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