Essay Example on A Title Is the height of the grass in my backyard normally Distributed

 Then I will calculate the expected values of the frequency table and compare them to the observed values that I have collected Next I will use the chi squared formula with a significance level of 5 comparing the critical value of my data found using significance level and degrees of freedom to my chi squared value If the chi squared calculated value is less than the critical value I will accept the null hypothesis that my data follows a normal distribution Criterion B I collected 125 blades of grass from an untrimmed patch in my backyard I used scissors to cut the blades as close to the ground as I could I then taped each blade flat on a piece of paper and measured their heights in centimeters starting from the bottom where they had been cut up to their tallest point The heights of the 125 blades of grass are as follows measured in centimeters to the nearest millimeter I then grouped my data into an interval frequency table and using 

This is again very near to the 99 74 that the amount of data that falls into this interval should be Using the chi squared test of independence χ 2 at a significance level of 5 I will determine if my data follows a normal distribution To find the expected values of my data I will use the use the normCDF function of my TI nspire for each height interval of the frequency table entering its lower and upper bounds the mean height of the total data 25 704 and the standard deviation 4 853 This function will give me a decimal representing the probability of data that falls within the selected height interval I will multiply this given decimal by 125 total blades collected giving me the expected value which I will round to the nearest tenth In the interval for example I will enter 20 for the lower bound 21 for the upper bound 25 704 for the mean and 4 853 for the standard deviation The format of this in the TI nspire is normCDF 20 21 25 704 4 853 0 046272 0 046272 125 an expected value of 5 78 for the interval the data follows a normal distribution with mean 25 704 cm and standard deviation 4 853 cm the data does not follow a normal distribution with mean 25 704 cm and standard deviation 4 853 cm Because I calculated the mean and standard deviation of my data the degrees of freedom is n 3 instead of n 1 So there are 11 degrees of freedom The critical value of 11 degrees of freedom with a significance level of 5 is 19 675 The chi squared formula is 13 1733 is less than 19 675 so I accept the null hypothesis