Overview
This video provides a comprehensive review of the normal distribution and the Chi-Squared test for independence, using basketball players' weights and performance data as practical examples.
Normal Distribution Concepts
- Given Data: Weight (W) of basketball players is normally distributed with mean (μ) = 66 kg and standard deviation (σ) = 4 kg.
- Probability Calculation:
- Find P(X < 60): Using the normal cumulative distribution function (N CDF), the probability that a player weighs less than 60 kg is approximately 0.668.
- Expected Number of Players Below 60 kg: In a group of 50 players, expected count = 50 * 0.668 = 3.34 players.
Probability Within 1.5 Standard Deviations
- Calculate the range: 1.5 * 4 = 6 kg.
- Interval: 66 - 6 = 60 kg to 66 + 6 = 72 kg.
- Probability P(60 < X < 72) = 0.866 using N CDF.
Finding a Weight Threshold Using Inverse Normal
- Given P(W > K) = 0.32, find K.
- Calculate P(W < K) = 1 - 0.32 = 0.68.
- Using inverse normal function, K ≈ 67.9 kg.
Chi-Squared Test for Independence
- Context: Testing if basketball players' performance is independent of their weight category.
- Hypotheses:
- H0: Performance is independent of weight.
- H1: Performance depends on weight.
- Expected Frequency Calculation:
- Example: For average weight and satisfactory performance, Expected frequency = (Total average weight * Total satisfactory performance) / Total players = (22 * 25) / 60 = 9.17.
Conducting the Chi-Squared Test
- Data entered into a 3x2 matrix.
- Chi-Squared statistic calculated as 2.49.
- Critical value at 5% significance level is 5.991.
- Since 2.49 < 5.991, fail to reject H0.
Conclusion
- There is insufficient evidence to conclude that performance depends on weight.
- At the 5% significance level, performance is independent of weight.
This video effectively demonstrates how to apply normal distribution calculations and Chi-Squared tests in real-world sports data analysis, providing clear steps and interpretations for statistical decision-making.
For a deeper understanding of the concepts discussed, you may find the following resources helpful:
- Comprehensive Review of Discrete Probability Distributions and Expected Values
- Understanding Z-Scores and their Applications in Statistics
- Mastering Confidence Intervals for Population Proportions in Statistics
- Introduction to Probability and Statistics: Key Concepts and Terminology
- Understanding and Calculating Standard Deviation in Science Labs
this video is about review for normal distribution and a Kai squared test the weight w of basketball players in a
tournament is found to be normally distributed with a mean of 66 and a standard deviation of 4 Kil gr find the
probability that a basketball player has a weight that is less than 60 kilg x follows a normal distribution
with a mean 66 standard deviation
4 we are looking for probability of X less than 60 since for normal distribution
ution it's a continuous data X less than 60 means X less than or equal to 60 do not be confused with the binomial
distribution we will use the N CDF menu 55
2 lower bond is negative Infinity upper bond is 60
mean 66 standard deviation four enter enter answer is a
0.668 A2 in a training section there are 50 basketball players find the expected number of players with a weight less
than 60 kilogram in this training session expect the number equals n * p 50 *
0.668 equals 3.34 the probability that a basketball
player has a weight that is within 1.5 or standard deviation of the mean is a q sketch a normal curve to represent this
probability standard deviation is four 1.5 * 4 = 6 mean is 66 - 6 = 60 66 + 6 = 72 let's
scale to the the graph mean 66 three equal segment on the left side and also on the right side the standard
deviation is a 4 70 74 78 minus 4
62 58 54 then sketch this uh belt curve we are
looking for 60 and a 72 60 is a here
72 estimator here then shate this area B2 find the value of Q we are
looking for probability of for X greater than 60 less than 72 we got to use the normal
CDF manual 55 2
60 72 66 4 enter enter answer is 0.866
say given that p of w greater than K = 0.32 find the value of K we will use inverse Norm for inverse
nor go to calculator Manu 553 we need a
area this area is defined by the area under the curve when X less than x in other words the area equals a
probability of for X less than x we are given probability of for w greater than k equal
0.32 so the area we need equals the probability of w less than k equal 1 minus probability of w greater
than K = 1 - 0.32 which is 0.68
0.68 Mu is 66 standard deviation is a four enter enter answer is 67.9 we used the this uh inverse
normal in calculator we got k equal 67.9 a basketball coach observed 60 of
her players to determine whether their performance and their weight were independent of each other her
observations were recorded as shown in the table she decided to conduct a High
Square test for Independence at the 5% significance level the critical value for this test
is 5991 for this test State the hypothesis independent of for is
ho dependent on is H1 because h is related to eal side to no change to no
relationship ho the performance is independent of the weight H1 the performance is dependent on the
weight D2 find the expected frequency of average weight we with the satisfactory
performance average weight and satisfactory performance this is a seven we need to figure out the
total number for average weight 22 total number for settis the factory
performance six + 7 + 12 25 so total for average is a
22 total for settis factory is a 25 total player 60
therefore 25 * 22 over 16
equals 9.17 e find the kai squar the statistic and a state with the
justification the conclusion for this test we need a matrix so clear this scratch Pad
menu seven enter enter we need uh three rows two
columns double click this uh right arrow button then go down right arrow button again go down enter put in the data then
store this Matrix into a control variables a
enter Then go to menu statistics seven stats
tests Kai Square 2E test eight right arrow button
eight enter we got Kai Square P value and D degree of
Freedom remember this please reject
H which means P value less than significance level reject H it's a
opposite Kai Square GC reject ho that's how you remember
when k Square greater than critical value K reject it all Kai square is a 2.49 this uh critical value is
5991 less than so not enough evidence reject ho we will write down ho with the significance level the performance is
independent of the weight at 5% significance level conclusion the performance is independent of the weight
at 5% significance level
Heads up!
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