The data below shows the heights in inches of 10 students in a class. StudentHeight, in inches Student 153 Student 252.5 Student
354 Student 451 Student 550.5 Student 649.5 Student 748 Student 853 Student 952 Student 1050 The standard error of the sample mean for this set of data is __________. Answer choices are rounded to the hundredths place. 0.59 1.87 1.77 0.19
1 answer:
Question not correct, so i have attached the correct question.
Answer:
SE = 0.59
Step-by-step explanation:
The mean of the students height is;
x' = (53 + 52.5 + 54 + 51 + 50.5 + 49.5 + 48 + 53 + 52 + 50)/10
x' = 51.35
Now, deviation from the mean for each height;
53 - 51.35 = 1.65
52.5 - 51.35 = 1.15
54 - 51.35 = 2.65
51 - 51.35 = -0.35
50.5 - 51.35 = -0.85
49.5 - 51.35 = -1.85
48 - 51.35 = -3.35
53 - 51.35 = 1.65
52 - 51.35 = 0.65
50 - 51.35 = -1.35
Now, square of the deviations above;
1.65² = 2.7225
1.15² = 1.3225
2.65² = 7.0225
-0.35² = 0.1225
-0.85² = 0.7225
-1.85² = 3.4225
-3.35² = 11.2225
1.65² = 2.7225
0.65² = 0.4225
-1.35² = 1.8225
Sum of the squared deviations;
2.7225 + 1.3225 + 7.0225 + 0.1225 + 0.7225 + 3.4225 + 11.2225 + 2.7225 + 0.4225 + 1.8225 = 31.525
Let's divide the sum by the DF of n - 1 i.e 10 - 1 = 9.
Thus;
31.525/9 = 3.50278
Taking the square root of that gives us the standard deviation.
Thus;
s = √3.50278
s = 1.8716
Formula for standard error is;
SE = s/√n
SE = 1.8716/√10
SE = 0.59
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