Please help me answer this TWO questions QUESTION 2 Homework: Chapter 13 Homewor

Question

Please help me answer this TWO questions

QUESTION 2

Homework: Chapter 13 Homework Save Score: 0 of 1 pt 1 of 2 (0 complete) HW Score: 0%, 0 of 2 pts 13.2.17-T Question Help * The billing department of a national cable service company is conducting a study of how customers pay their monthly cable bills. The cable company accepts payment in person at a local office, by mail, by credit card, or by electronic funds transfer from a bank account. The cable company randomly sampled 400 customers to determine if there is a relationship between the customer's age and the payment method used. The accompanying sample results were obtained. Based on the sample data, can the cable company conclude that there is a relationship between the age of the customer and the payment method used? Conduct the appropriate test at the = 0.01 level of significance Click the icon to view the sample data Chi-square test of independence Chi-square goodness of fit test One-way analysis of variance Two sample t-test Determine the null and alternative hypotheses. Choose the correct answer below A. Ho. Customer's age is normally distributed. OB. Hg: The mean customer age is equal to the mean payment method. C. Hg: The mean age for each payment method is equal. HA: Customer's age is not normally distributed. HA: The mean customer age is not equal to the mean payment method. HA: At least one mean for a payment method is not equal to the others. HA: Customer's age is not independent of payment method D. Ho: Customer's age is independent of payment method. Calculate the test statistic. The test statistic is(Round to two decimal places as needed.)

Explanation / Answer

Q2.

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, = 0.1

from standard normal table, chi square value at right tailed, ^2 /2 =4.605

since our test is right tailed,reject Ho when ^2 o > 4.605

we use test statistic ^2 o = (Oi-Ei)^2/Ei

from the table , ^2 o = 114.469

critical value

the value of |^2 | at los 0.1 with d.f (r-1)(c-1)= ( 3 -1 ) * ( 2 - 1 ) = 2 * 1 = 2 is 4.605

we got | ^2| =114.469 & | ^2 | =4.605

make decision

hence value of | ^2 o | > | ^2 | and here we reject Ho

^2 p_value =0

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: 114.469

critical value: 4.605

p-value:0

decision: reject Ho

MATRIX col1 col2 TOTALS row 1 53 215 268 row 2 153 176 329 row 3 245 151 396 TOTALS 451 542 N = 993

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