Statement of Tasks: The plan is to obtain data of female juniors involved in sports and the relevance to their grade point averages. The data will be collected through a survey. The survey will be distributed to fifty female juniors that attend Taft High School. The survey will ask the student to state the curriculum such as IB or other (Regular/Honors). Students will also have to state their grade point averages. The researcher has to assure that each student is either involved or has recently been involved in a sport. A statistical analysis will be applied to the data in order to uncover any relevance of the data. The analysis will require a chi-squared (x2) test.
Detailed Plan: I, the researcher, surveyed fifty female juniors
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I add up all the GPAs and then divide by the total number of GPAs.
5.1+ 4.6+ 4.7+ 4.7 + 4.7+ 4.8 + 4.3 + 4.2 + 4.6 + 4.3 + 4.3+ 4.6 + 4.8 + 4.5+ 3.8 + 3.6 + 3.8+ 3.5 + 3.3 + 3.3 = 85.5
85.5/20=4.28
= 4.28
This is the mean found for Others GPA.
4.1 + 3.6 + 3.4 + 3.4 + 3.65 + 3.48 + 3.2 + 3.8 + 3.7 + 3.8 + 3.1+ 3.9 + 3.2 + 3.1+ 3.2 + 3.0+ 3.0 + 2.9+ 2.9 + 2.5+ 3.0+ 2.7+ 3.0 + 2.8+ 2.7 + 2.6 + 2.0+ 2.0 + 1.7+ 2.0= 92.43
92.43/30=3.08
= 3.08
In order to find the median for IB and Others GPA, I must put the GPAs in numerical order. Since both totals are even numerals, I have to take the mean of the two middle numbers of the set of GPA. IB median: 4.3+4.5= 8.8 8.8/2= 4.4
Others median: 3.0+ 3.1=6.1 6.1/2 = 3.05
There several modes within each data set. I simply look for repeating digits. Since there is a handful, I will only choose of the many modes to be the main mode for the sets.
IB mode: 4.7 and 4.3
Others mode: 3.0
This is some of the input of GPAs in a graphic display calculator (GDC). I will use the GDC to show the central tendency. First, I input the GPA. After, the data was in I pressed on STAT.
Then, I went to CALC and pressed ENTER on the first option.
1-Var stats (IB)
1-Var Stats (Others)
x2 test for independence
≤ 5.0 ≤ 4.0 ≤ 3.0 Total
IB 14 6 0 20
Other 1 14 15 30
Total 15 20 15 50
Observed frequency (Fo)
1) First, I must set up
Indicating the individual number 65 gives a 5 point range to the mean. It seems the median is the most accurate way to discribe the data set, as it uneffected by the outlier value.
It is one of the most popular and well known measures of central tendency. It can be used with both discrete and continuous data, although its use is most often with continuous data
2. Based on the scale of measurement for each variable listed below, which measure of central tendency is most appropriate for describing the data?
+ (5% are 450 sec) + (3% are 125 sec) + (2% are 120 sec)
For the education data, I must say to calculate the median, from high school to college graduate, the median would be the some college portion since it would be the middle. The median for the experimental group would be 11 and 34.4% and for the control group it would be 15 and 41.7%.
1. By hand, compute the mean, median, and mode for the following set of 40 reading scores:
Number of patients in clinic = 4 + 5 + 7.73 + 9.86 + 0.4 + 0.6 + 5 + 6.6
2.34 0.58 71.03 5.31 112.33 51.17 118.39 121.45 6.66 1.23 0.70 0.61 6.87 1.67 42.37% 18.46% 10.25% 14.81% 41.85%
5,016 * .35 = 1,755.6 * 6 = 10,533.6 5,517.6 * .35 = 1,931.16 * 6 = 11,586.96
665 0.34 174 216 537 234 122 1.2 0.2 60.4 9.6 0.22 65.0 2,149 554 66.6 2,483 80,300 492 14 23 6 21 15 874 524 12,216 30
2) The salaries of ten randomly selected physicians are shown below. Find the median salary.
3.375(1 + r)-1 + 3.375 (1 + r)-2 + 3.375 (1 + r)-3 + ...…+ 3.375 (1 + r)-40+100(1 + r)-40 = 95.6
= (89.14 + 83.72 + 82.7 + 86.54 + 93.87 + 88.88 + 92.5 + 90 + 97.5) ÷ 9
Min 3.2x1 + 2.2x2 + 4.2x3 + 3.9x4 + 1.2x5 + 0.3x6 + 2.1x7 + 3.1x8 + 4.4x9 + 2.7x10 + 4.7x11 + 3.4x12 + 2.1x13 + 2.5x14 + 6.0x15 + 5.2x16 + 5.4x17 + 4.5x18 + 6.0x19 + 3.3x20 + 2.7x21 + 5.4x22 + 3.3x23 + 2.4x24 + 0.3x25 +0.7x26 + 3.5x27
26.0 $(9,240 + 95,000 + 5,400 + 26,180 + 109.91 + 21,600 + 12,000 + 9,000 + 11,200 + 8,000 + 15,000)