00:01
In this question we have given that a professor done a survey and in survey he observed that he predict that 60 % of students purchase a book, 25 % of the students print it out from web and 15 % of students will read it online.
00:45
Here we have given the total number of students is equal to 126.
00:59
Out of this, 71 bought the book, 30 says that they will print the book and 25 says that they will read online.
01:26
So using this information, first we have to state the hypothesis for the testing if professor's prediction were inaccurate.
01:36
So the prediction were inaccurate when the hypothesis are edge not that is null hypothesis probability of buying book on book will be 0 .6 probability of print will be given as 0 .25 the probability of read online is 0 .50 and the alternative hypothesis will be ha that is at least of the claimed probability is 0 so this is the hypothesis that are inaccurate so out of the given option option c is correct so this is answer for the first question.
02:51
Now in part b we have to find the expected values.
03:01
So for that we will consider a table that is as follows.
03:06
So here we have the category the observed value that is oi, the expected value that can be calculated using the form of ei is equal to total multiplied with the probability.
03:37
So here we have the first category that is they bought the book.
03:42
The observed number is 71 and the expected value will be total we have 126 students and out of 126 60 % have purchased book so it will be equals to 75 .6.
03:58
Now the category 2 that is they print the book.
04:02
So for that we have given the observed value.
04:04
Is 30.
04:06
So expected value will be total that is 126 and the probability that is given to us is 25%.
04:18
So if we multiply these two terms then we'll get it as 31 .5.
04:23
Now the third category is they read online.
04:26
So for that we have given the observed value as 25 and the expected value will be 126 multiplied with 15%.
04:36
So this is equal to 8%.
04:38
18 .9.
04:39
So these are the expected value for the given category.
04:44
So we'll conclude answer for this part that is as follows.
04:56
By hard copy, the expected value will be equals to 75 .6.
05:09
Print out.
05:12
So the expected value will be 31 .5 and for the case, that is read online the expected value is equal to 18 .9...