In: Statistics and Probability
2. A publishing house conducted a survey to assess the reading habits of teenagers. The company publishes four types of books specifically tailored to suit the interests of teenagers. Management hypothesizes that there are no differences in the preferences for the four types of books. A sample of 1400 teenagers indicated the following preferences for the four book types.
Publication Frequency of Preference
Non-Fiction 445
Mystery 515
Humor 150
Short Stories 290
TOTAL 1400
Management needs your expertise to determine whether there are differences in preferences for the four types of books. Hint: This problem is the same format as the M&Ms example discussed in class.
a. State the null and alternate hypotheses in plain English sentences (not equations).
b. What is the calculated chi-squared value? Show your calculations.
GIVEN THAT :-
According to the question we have that ,
There are 4 types of publications and the frequency of preference by the teenagers gang of 1800 people
now to find
TO FIND :-a) state the null and alternate hypothesis .
NULL HYPOTHESIS :- It is denoted or represented as H0 . this is defined a there is no difference in preference of the four types of books.
ALTERNATIVE HYPOTHESIS:-It is denoted or represented with H1. this is defined as there is difference in the books.
TO FIND:-b) what is the calculated chi-squared value.
here at first taking the
expected preference as ei
frequency of preference as fi
expected frequency = sum of four frequencies / no of frequencies
=1400/4
ei =350.
now finding the (fi-ei)^2/ei
so for non friction = 25.78
mystry = 77.78
humor = 114.285
short storys = 48.28
now the test statistic is X^2= sum of (fi-ei)^2/ei
X^2= 266.125
for finding the degree of freedom we have a formula is n-1
from the above we have n=4
degree of freedom = 4-1 = 3 .
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