A credit card watchdog group claims that there is a difference in the mean credi

Question

A credit card watchdog group claims that there is a difference in the mean credit card debts of households in New York and Texas. The results of a random survey of 250 households from each state are shown below New York Texas LaTeX: overline{x_1}x1¯¯¯¯¯x1¯ = $4446.25 LaTeX: overline{x_2}x2¯¯¯¯¯x2¯ = $4567.24 S1 =$1045.70 S2 = $1361.95 n1 = 250 n2 = 250 Set up the null and alternative hypotheses to check the watchdog group's claim. a) Ho: LaTeX: mu_1gemu_21212 LaTeX: vsvsvs Ha: LaTeX: mu_111LaTeX: <mu_2<2<2 b) Ho: LaTeX: mu_1:=:mu_2:vs:H_a:::mu_1 e:mu_21=2vsHa121=2vsHa12 c) LaTeX: H_o:mu_1le:mu_2:vs:H_a:mu_1>:mu_2Ho12vsHa1>2Ho12vsHa1>2 2. Find the standardized test statistic 3. what test form of rejection region do we have here?

Explanation / Answer

Given that,
mean(x)=4446.25
standard deviation , s.d1=1045.7
number(n1)=250
y(mean)=4567.24
standard deviation, s.d2 =1361.95
number(n2)=250
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, = 0.05
from standard normal table, two tailed t /2 =1.96
since our test is two-tailed
reject Ho, if to < -1.96 OR if to > 1.96
calculate pooled variance s^2= (n1-1*s1^2 + n2-1*s2^2 )/(n1+n2-2)
s^2 = (249*1093488.49 + 249*1854907.8025) / (500- 2 )
s^2 = 1474198.1463
we use test statistic (t) = (x-y)/sqrt(s^2(1/n1+1/n2))
to=4446.25-4567.24/sqrt((1474198.1463( 1 /250+ 1/250 ))
to=-120.99/108.5983
to=-1.1141
| to | =1.1141
critical value
the value of |t | with (n1+n2-2) i.e 498 d.f is 1.96
we got |to| = 1.1141 & | t | = 1.96
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value: two tailed ( double the one tail ) - ha : ( p != -1.1141 ) = 0.2658
hence value of p0.05 < 0.2658,here we do not reject Ho
ANSWERS
---------------
null, Ho: u1 = u2 , alternate, H1: u1 != u2
test statistic: -1.1141
reject Ho, if to < -1.96 OR if to > 1.96
critical value: -1.96 , 1.96
decision: do not reject Ho
p-value: 0.2658

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