Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

Consider the following competing hypotheses and accompanying sample data. Use Ta

ID: 3438015 • Letter: C

Question

Consider the following competing hypotheses and accompanying sample data. Use Table 1.

Calculate the value of the test statistic. (Negative values should be indicated by a minus sign.Round intermediate calculations to 4 decimal places and final answer to 2 decimal places.)

At the 5% significance level, calculate the critical value. (Negative value should be indicated by a minus sign. Round your answer to 3 decimal places.)

Consider the following competing hypotheses and accompanying sample data. Use Table 1.

H0: p1 p2 0 HA: p1 p2 < 0

Explanation / Answer

Null Hypothesis, Ho: p1 - p2 >=0
Alternate Hypothesis, H1: p1 - p2 <0
Test Statistic
Sample 1 : X1 =224, n1 =386, P1= X1/n1=0.58
Sample 2 : X2 =245, n2 =386, P2= X2/n2=0.635
Finding a P^ value For Proportion P^=(X1 + X2 ) / (n1+n2)
P^=0.608
Q^ Value For Proportion= 1-P^=0.392
we use Test Statistic (Z) = (P1-P2)/(P^Q^(1/n1+1/n2))
Zo =(0.58-0.635)/Sqrt((0.608*0.392(1/386+1/386))
Zo =-1.548
| Zo | =1.55
Critical Value
The Value of |Z | at LOS 0.05% is 1.645
We got |Zo| =1.548 & | Z | =1.645
Make Decision
Hence Value of |Zo | < | Z | and Here we Do not Reject Ho
P-Value: Left Tail - Ha : ( P < -1.5478 ) = 0.06083
Hence Value of P0.05 < 0.06083,Here We Do not Reject Ho


a) Zo =-1.55
b) The Value of |Z | at LOS 0.05% is -1.645
c) Do not reject H0 since the value of the test statistic is less than the critical value.

Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Drop an Email at
drjack9650@gmail.com
Chat Now And Get Quote