CHAPTER1 z. In 2008, A New York Times poll of Americans with at least a four yea
ID: 3314481 • Letter: C
Question
CHAPTER1 z. In 2008, A New York Times poll of Americans with at least a four year college degree asked them how they would rate their overall experience as an undergraduate student. T below. In 2016, a survey of 500 randomly selected graduates were surveyed on how they overall experience as an undergraduate student, the results of the survey are listed in the table. Determine if the ratings of distribution of college graduates in 2016 level of significance. (DO NOT ROUND hese results are listed in the table would rate their fits the initial distribution. Use a 5% YOUR CALCULATIONS- USE 4 DECIMAL PLACES) |DistributionI :| EXPECTED|-OE) | College )(0 OBSERVED: Graduates Distribution Co - E)2 Rating of of ratings in Distribution 2008 Sample EXPECTED(O-E) (0 O-EP overall experience Excellent in 2015 54% 259 Good 39% 192 39 10 500 Fair 6% Poor 1% 100% 500 0 (a) State the null and alternate hypotheses and the level of significance b) Find the critical valuefs). Sketch the sampling distribution and show the area corresponding to the critical value (c) Calculate the sample test statistic from the table above. (d) Based on your answers for parts (a) through (c), will you reject or fail to reject the null hypothesis? Interpret your conclusion in the context of the application
Explanation / Answer
ChiSquare Test For INDEPENDENCE TEST
observed frequencies are Oi
259 192 39 10
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expected frequencies are Ei
0.54*500 = 270, 0.39*500 = 195, 0.06*500 = 30, 0.01*500 = 5
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and claiming hypothesis is
null, Ho: no association exists b/w them OR observations are independent
alternative, H1: exists association b/w them OR observations are dependnet
level of significance, = 0.05
from standard normal table, chi square value at right tailed, ^2 /2 =5.9915
since our test is right tailed,reject Ho when ^2 o > 5.9915
we use test statistic ^2 o = (Oi-Ei)^2/Ei
from the table take sum of (Oi-Ei)^2/Ei, we get ^2 o = 8.1943
critical value
the value of |^2 | at los 0.05 with d.f, n - 1 = 3 - 1 = 2 is 5.9915
we got | ^2| =8.1943 & | ^2 | =5.9915
make decision
hence value of | ^2 o | > | ^2 | and here we reject Ho
^2 p_value =0.0166
ANSWERS
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null, Ho: no significance between the how they would rate between overall experience
alternative, H1: significance between the how they would rate between overall experience
test statistic: 8.1943
critical value: 5.9915
p-value:0.0166
decision: reject Ho
conclude that there is significance between the how they would rate between overall experience
Observed (Oi ) Expected ( Ei) Oi-Ei (Oi-Ei)^2 (Oi-Ei)^2/Ei 259 270 -11 121 0.4481 192 195 -3 9 0.0462 39 30 9 81 2.7 10 5 5 25 5