Answer:
The claim that the scores of UT students are less than the US average is wrong
Step-by-step explanation:
Given : Sample size = 64
Standard deviation = 112
Mean = 505
Average score = 477
To Find : Test the claim that the scores of UT students are less than the US average at the 0.05 level of significance.
Solution:
Sample size = 64
n > 30
So we will use z test

Formula : 


Refer the z table for p value
p value = 0.9772
α=0.05
p value > α
So, we accept the null hypothesis
Hence The claim that the scores of UT students are less than the US average is wrong
Answer:
11 ;
Step-by-step explanation:
Given the data:
30 27 22 25 24 25 24 15 35 35 33 52 49 10 27 18 20 23 24 25 30 24 24 24 18 20 25 27 24 32 13 13 21 2 37 35 32 33 29 3 28 28 25 29 31
Number of classes = 5
Class width : Range / number of classes
Class width = (maximum - minimum) / 5
Class width = (52 - 2) / 5 = 50/5 = 10 + 1 = 11
For the frequency table showing class limits, class boundaries, midpoints, frequencies, relative frequencies, and cumulative frequencies and histogram
Kindly check attached picture
Answer:
y-1=4/5(x-2/3)
Step-by-step explanation:
y-y1=m(x-x1)
y-1=4/5(x-2/3)
71/20 i believe. Is it multiple choice?