The following histogram shows the exam scores for a Prealgebra class. Use this histogram to answer the questions.Prealgebra Exam
ScoresScores 70.5, 75.5, 80.5, 85.5, 90.5, 95.5, 100.5Frequency 0, 4, 8, 12, 16, 20, 24Step 1 of 5:Find the number of the class containing the largest number of exam scores (1, 2, 3, 4, 5, or 6).Step 2 of 5:Find the upper class limit of the third class.Step 3 of 5:Find the class width for this histogram.Step 4 of 5:Find the number of students that took this exam.Step 5 of 5:Find the percentage of students that scored higher than 95.595.5. Round your answer to the nearest percent.
2 answers:
Answer:
The number of the class containing the largest score can be found in frequency 24 and the class is 98 - 103
For the third class 78 - 83 ; the upper limit = 83
The class width for this histogram 5
The number of students that took the exam simply refers to the frequency is 84
The percentage of students that scored higher than 95.5 is 53%
Step-by-step explanation:
The objective of this question is to use the following histogram that shows the exam scores for a Pre-algebra class to answer the question given:
NOW;
The table given in the question can be illustrated as follows:
S/N Class Score Frequency
1 68 - 73 70.5 0
2 73 - 78 75.5 4
3 78 - 83 80.5 8
4 83 - 88 85.5 12
5 88 - 93 90.5 16
6 93 - 98 95.5 20
<u>7 98 - 103 100.5 24 </u>
TOTAL: 84
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a) The number of the class containing the largest score can be found in frequency 24 and the class is 98 - 103
b) For the third class 78 - 83 ; the upper limit = 83 ( since the upper limit is derived by addition of 5 to the last number showing in the highest value specified by the number in the class interval which is 78 ( i.e 78 + 5 = 83))
c) The class width for this histogram 5 ; since it is the difference between the upper and lower boundaries limit of the given class.
So , from above the difference in any of the class will definitely result into 5
d) The number of students that took the exam simply refers to the frequency ; which is (0+4+8+12+16+20+24) = 84
e) Lastly; the percentage of students that scored higher than 95.5 is ;
⇒
= 0.5238095 × 100
= 52.83
To the nearest percentage ;the percentage of students that scored higher than 95.5 is 53%
Answer:
1. 98-103 (6th class)
2. 88
3. 5
4. 84
5. 52%
Step-by-step explanation:
Find attached the frequency table.
The class of exam scores falls between (1, 2, 3, 4, 5, or 6).
The exam score ranged from 68-103
1) The largest number of exam scores = 24
The largest number of exam scores is in the 6th class = 98 -103
Step 2 of 5:
The upper class limit is the higher number in an interval. Third class interval is 83-88
The upper class limit of the third class 88.
Step 3 of 5:
Class width = upper class limit - lower class limit
We can use any of the class interval to find this as the answer will be the same. Using the interval between 73-78
Class width = 78 - 73
Class width for the histogram = 5
Step 4 of 5:
The total of students that took the test = sum of all the frequency
= 0+4+8+12+16+20+24 = 84
The total of students that took the test = 84
Step 5 of 5:Find the percentage of students that scored higher than 95.5
Number of student that scored higher than 95.5 = 20 + 24 = 44
Percentage of students that scored higher than 95.5 = [(Number of student that scored higher than 95.5)/(total number of students that took the test)] × 100
= (44/84) × 100 = 0.5238 × 100 = 52.38%
Percentage of students that scored higher than 95.5 = 52% (nearest percent)
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Answer:
e. No, the p-value is greater than 0.01, so there is insufficient evidence at the 0.01 level to conclude that pain levels are lower after hypnotism.
Step-by-step explanation:
Hello!
The study's objective is to test if hypnotism reduces pain.
A sample of n=8 subjects was taken and the pain level was recorded in each subject before and after being hypnotized.
The variable of interest was determined by calculating the difference of pain level after - before hypnosis. This is a paired sample test and the variable can be determined as:
Xd: Difference between pain level felt after hypnosis and pain level felt before hypnosis of a subject.
The sample average and standard deviation obtained were:
Xd= -3
Sd= 3
And the variable is presumed to be approximately normal.
An approximately normal distribution is enough to conduct a paired sample t-test.
If the claim is that hypnosis reduces pain, then the average pain level after hypnosis should be less than the average pain level before hypnosis, then the average difference is expected to be negative, symbolically: μd < 0
The test will be one-tailed and so will be the p-value.
Remember: The p-value is defined as the probability corresponding to the calculated statistic if possible under the null hypothesis (i.e. the probability of obtaining a value as extreme as the value of the statistic under the null hypothesis).
So first step is to calculate the value of the statistic under the null hypothesis and then you can calculate the p-value.
H₀: μd ≥ 0
H₁: μd < 0
The DF of the t-test are n-1= 7
Then you can calculate the p-value as:
P(t₇≤-2.83)= 0.0127
The level of the test is α: 0.01
The decision rule is:
If p-value ≤ α, reject the null hypothesis.
If p-value > α, do not reject the null hypothesis.
The p-value > α the decision is to not reject the null hypothesis.
Correct option: e. No, the p-value is greater than 0.01, so there is insufficient evidence at the 0.01 level to conclude that pain levels are lower after hypnotism.
I hope this helps!