The National Sleep Foundation used a survey to determine whether hours of sleepi

Question

The National Sleep Foundation used a survey to determine whether hours of sleeping per night are independent of age (Newsweek, January 19, 2004). The following show the hours of sleep on weeknights for a sample of individuals age 49 and younger and for a sample of individuals age 50 and older. Hours of Sleep Age Fewer than 6 6 to 6.9 7 to 7.9 8 or more Total 49 or younger 30 64 74 72 240 50 or older 32 85 260 a. Conduct a test of independence to determine whether the hours of sleep on weeknights are independent of age. Use a = .05. Use Table 12.4. Compute the value of the X? test statistic (to 2 decimals). .47 3 The p value is greater than .109 What is your conclusion? Cannot reject the assumption that age and hours of sleep are independent o b. Using the total sample of 500, estimate the percentage of people who sleep less than 6,6 to 6.9, 7 to 7.9, and 8 or more hours on weeknights (to 1 decimal). Less than 6 hours 13.8 8 % 6 to 6.9 hours | 24.4 8% 7 to 7.9 hours 28.6 % 8 or more hours (33.2 % "Hide Feedback Partially Correct Check My Work

Explanation / Answer

a.

b. total sample = 500

percentage of people sleep less than 6 hours = 62/500 = 12.4%
percentage of people sleep 6 to 6.9 hours =127/500 = 25.4%
percentage of people sleep 7 to 7.9 hours = 154/500 = 30.8%
percentage of people sleep 8 or more = 157/500 = 31.4%

Given table data is as below MATRIX col1 col2 col3 col4 TOTALS row 1 30 64 74 72 240 row 2 32 63 80 85 260 TOTALS 62 127 154 157 N = 500 ------------------------------------------------------------------

calculation formula for E table matrix E-TABLE col1 col2 col3 col4 row 1 row1*col1/N row1*col2/N row1*col3/N row1*col4/N row 2 row2*col1/N row2*col2/N row2*col3/N row2*col4/N ------------------------------------------------------------------

expected frequecies calculated by applying E - table matrix formulae E-TABLE col1 col2 col3 col4 row 1 29.76 60.96 73.92 75.36 row 2 32.24 66.04 80.08 81.64 ------------------------------------------------------------------

calculate chisquare test statistic using given observed frequencies, calculated expected frequencies from above Oi Ei Oi-Ei (Oi-Ei)^2 (Oi-Ei)^2/Ei 30 29.76 0.24 0.06 0 64 60.96 3.04 9.24 0.15 74 73.92 0.08 0.01 0 72 75.36 -3.36 11.29 0.15 32 32.24 -0.24 0.06 0 63 66.04 -3.04 9.24 0.14 80 80.08 -0.08 0.01 0 85 81.64 3.36 11.29 0.14 ^2 o = 0.58 ------------------------------------------------------------------

set up null vs alternative as
null, Ho: no relation b/w X and Y OR X and Y are independent
alternative, H1: exists a relation b/w X and Y OR X and Y are dependent
level of significance, = 0.05
from standard normal table, chi square value at right tailed, ^2 /2 =7.81
since our test is right tailed,reject Ho when ^2 o > 7.81
we use test statistic ^2 o = (Oi-Ei)^2/Ei
from the table , ^2 o = 0.58
critical value
the value of |^2 | at los 0.05 with d.f (r-1)(c-1)= ( 2 -1 ) * ( 4 - 1 ) = 1 * 3 = 3 is 7.81
we got | ^2| =0.58 & | ^2 | =7.81
make decision
hence value of | ^2 o | < | ^2 | and here we do not reject Ho
^2 p_value =0.9


ANSWERS
---------------
null, Ho: no relation b/w X and Y OR X and Y are independent
alternative, H1: exists a relation b/w X and Y OR X and Y are dependent
test statistic: 0.58
critical value: 7.81
p-value:0.9
decision: do not reject Ho

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