Apply the Friedman test to the accompanying table of data to determine whether we can conclude t

Apply the Friedman test to the accompanying table of data to...

Key Concepts

Significance Level (?)

A preset threshold (in this case, 0.10) that determines the criteria for rejecting the null hypothesis. It represents the probability of a Type I error, which is rejecting the null hypothesis when it is actually true.

Rank-Based Analysis

This approach converts raw data into ranks, reducing the impact of outliers and non-normal distributions. In the Friedman test, ranking is used to compare the relative performance of treatments across different blocks or subjects.

Hypothesis Testing

A framework used to assess whether observed data can be attributed to chance under a null hypothesis. For the Friedman test, the null hypothesis typically states that there are no differences among treatments, and the alternative hypothesis suggests that at least two treatments differ.

Nonparametric Methods

These methods do not assume a specific probability distribution for the data, making them suitable for ordinal data or when the assumptions of parametric tests (like normality) are violated. The Friedman test is an example of a nonparametric method.

Friedman Test

A nonparametric statistical test used to detect differences in treatments across multiple test attempts. It analyzes ranked data from repeated measures or matched subjects to determine if there is a significant difference between group medians without assuming a normal distribution.

Repeated Measures or Block Designs

This concept involves collecting data from the same subjects or experimental units under different conditions or at different times. This design helps control for variability among subjects, and the Friedman test is specifically designed to analyze such data.

Transcript

00:01 Hello students we are given in the question assume that the population are normal distributed test where than mu 1 is greater than mu 2 at the time alpha equals to 0 .1 level of significance mean 90 % confidence interval construct a 90 % confidence interval about mu 1 minus mu 2 right so these are given us so here what we are given we are given for sample 1 and sample 2 right for sample 1 if i right here so sample one and one is basically which is our sample number so number of samples so basically this is 25 and sample size this is 18 and x1 bar is basically 50 .2 and here x2 bar so basically this is the this is the mean for the sample and this is 42 .0 and s one for sample one is standard evaluation for sample one and standard deviation for sample two right and mu is basically the mean for sample so basically this is the mean for sorry this is the mean for samples and mu one and mu two r mu one and mu two r means for for our population means right so if i write our population mean our population mean thus sigma and sigma 1 and sigma 2 are population standard deviation right so let's talk about the a part in the 8th part what we need to do is if i talk about a what we need to here if i see we need to test whether test whether mu 1 is greater than mu 2 mu 1 is greater than mu 2 basically mu 1 is not equal to mu 2 at l 5 alpha equals to point one level of significance right now if i write for this so first we consider our null hypothesis basically we consider mu 1 equals to mu 2 and then we goes up to mu 1 is not equal to mu 2 which have two pillars mu 1 is greater than mu 2 and mu 1 is less than mu 2 right so basically if this is not considered and when and then we reject our null hypothesis.

02:40 So let us come.

02:45 So now if i come here, so we have to the test static is test statistics.

02:59 So basically this is t0 is equals to x1 bar minus x2 bar minus muo 1 minus mu 2 divided by under root of s1 squared divided by n1 plus.

03:16 S2 squared divided by n2 right so basically first we have we have to consider null hypothesis so that if it is not consider then we will reject it or not so x1 minus x2 we need to consider first so what is our x1 minus x2 so this is 50 .2 minus 40 52 point sorry basically this is 50 .2 right so if i write for this this is basically 50 .2 right so if i write for this so this is basically 50 .2 minus if i come there so what what is my x2 bar value 42 .0 right mine divided by under root s1 square what is our s1 so as 1 is 64 square divided by 25 64 square divided by 25 plus plus if i write 64 square divided by 25 and 9 .9 .9 square divided by 18 right so basically this is this is 8 .2 divided by under root so under root 1 so if i calculate this roughly this will come out as 1 .6384 plus now calculate this this 99 .99, 9 .9, multiply by 9 .9 .98 .01 divided by 80, which is again comes out as 5 .445.

05:01 Now let us calculate my final value.

05:04 So basically this is 1 .6384 plus 5 .445.

05:12 So 7 .083 .4.

05:13 So 7 .083 .4.

05:16 Right so this is basically 7 .1 so 8 .2 under root 7 .1 right so 8 .2 divided by under root 7 .08 sorry 7 .08 34 so basically this is 3 .08 8 .08 so basically this is 3 .08 right 3 .08 so t not comes at 3 .08, right? now let us calculate other values.

05:53 We need to find p value in both cases when sample size is basically we have given for our first case and second case, right? so which we need to test 25 and 18...

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