Perform a test of the hypothesis TABLE 1: Complete Contingency Table TABLE 2: Su
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Question
Perform a test of the hypothesis
TABLE 1: Complete Contingency Table
TABLE 2: Summary of the Hypothesis Test
According to the "random-walk model" for stock prices, the price movements of stocks on a given day are not influenced by movements in the prices on previous days. We've recorded closing price information for hundreds of NASDAQ stocks from last Monday, Tuesday, and Wednesday with the aim of testing this model. Table 1 below, a contingency table, displays some information for a random sample of 500 stocks.
The table summarizes the closing price information of the stocks relative to the closing prices of the day before. Each cell of the table has three numbers: the first number is the observed cell frequency (f sub 0); the second number is the expected cell frequency (f sub E); under the assumption that there is no association between the two variables closing price movement from Monday to Tuesday and closing price movement from Tuesday to Wednesday; and the third number is the value
(SEE FORMULA in uploaded image.)
Fill in the missing values of Table 1. Round your expected frequencies to at least two decimal places, and round your values to at least three decimal places. Then, using the 0.05 level of significance, perform a test of the hypothesis that there is no association between the two variables closing price movement from Monday to Tuesday and closing price movement from Tuesday to Wednesday. Complete Table 2 based on your test.
GIVEN:
random sample of 500 stocks.
0.05 level of significance
COMPLETE TABLE 1: Complete Contingency Table
COMPLETE TABLE 2: Summary of the Hypothesis Test
and
ANSWER QUESTIONS
1. Type of test statistic: drop down choices are: Z, t, chi Square, F
2. Value of the test statistic. (Round your answer to at least two decimal places).
3. critical value for a test at the 0.05 level of significance. (Round your answer to at least two decimal places).
4. Can we conclude that there is an association between the variables closing price movement from Monday to Tuesday and closing price momentum from Tuesday to Wednesday. Use the 0.05 level of significance. YES or NO
According to the "random walk model" for stock prices, the price movements of stocks on a given day are not influenced by movements in the prices on previous days. We've recorded closing price information for hundreds of NASDAQ stocks from last Monday, Tuesday, and Wednesday with the aim of testing this model. Table 1 below, a contingency table, displays some information for a random sample of 500 stocks. The table summarizes the closing price information of the stocks relative to the closing prices of the day before. Each cell of the table has three numbers: the first number is the observed cell frequency (fo); the second number is the expected cell frequency (fa) under the assumption that there is no association between the two variables closing price movement from Monday to Tuesday and closing price movement from Tuesday to Wednesday, and the third number is the value (fo-fa) (observed cell frequency Expected cell frequency)2 Expected cell frequency The numbers labeled "Total" are totals for observed frequency. (fo-fa) Fill in the missing values of Table 1. Round your expected frequencies to at least two decimal places, and round your values to at least three decimal places. Then, using the 0.05 evel of significance, perform a test of the hypothesis that there is no association between the two variables closing price movement from Monday to Tuesday and closing price movement from Tuesday to Wednesday. Complete Table 2 based on your test. Table 1 Contingency TableExplanation / Answer
Given table data is as below
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calculation formula for E table matrix
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expected frequecies calculated by applying E - table matrix formulae
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calculate chisquare test statistic using given observed frequencies, calculated expected frequencies from above
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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 =3.8415
since our test is right tailed,reject Ho when ^2 o > 3.8415
we use test statistic ^2 o = (Oi-Ei)^2/Ei
from the table , ^2 o = 5.1807
critical value
the value of |^2 | at los 0.05 with d.f (r-1)(c-1)= ( 2 -1 ) * ( 2 - 1 ) = 1 * 1 = 1 is 3.8415
we got | ^2| =5.1807 & | ^2 | =3.8415
make decision
hence value of | ^2 o | > | ^2 | and here we reject Ho
^2 p_value =0.0228
ANSWERS
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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: 5.1807
critical value: 3.8415
p-value:0.0228
decision: reject Ho
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