Question

Reserve Problems Chapter 10 Section 1 Problem 1 Consider the hypothesis test Ho : M1 - H2 = 0) against Hi : M1 - H2 + 0 samples below: I 35 39 32 32 33 30 31 30 39 38 31 37 37 32 39 32 35 40 II 34 30 33 32 31 29 30 38 31 33 30 30 29 33 34 34 Variances: 01 = 3.7, 62 = 1.8. Use a = 0.05. (a) Test the hypothesis and find the P-value. Find the test statistic. Round your answers to four decimal places (e.g. 98.7654). ZO P – value = (b) Explain how this test could be conducted with a confidence interval. Round your answers to three decimal places (e.g. 98.765). sui – 425 We are 95% confident that value 1 exceeds value 2 by between (c) What is the power of the test for the true difference in means of 2? Round your answer to three decimal places (e.g. 98.765). The power of the test is 1-B=

Answer 1

We can find the sample mean of the both data set using excel function =AVERAGE( ) and =STDEV.S() respectively.

B D E TI А Hourly 1 Salaried 2 35 34 3 39 30 4 32 33 5 32 =AVERAGE(A2:A19) 34.5556 32 =AVERAGE(D2:017) 31.9375 31 6 33 7 30 29 8 31 30 9 30 38 31 33 30 30 29 10 39 11 38 12 31 13 37 14 37 15 32 16 39 17 32 18 35 19 40 33 34 34

Но

11
-
112
= 0 vs $1596087686770_blob.png$
11
-
112
≠ 0   

Given :
1
= 34.5556  ,
01
= 3.7, n₁ = 18 and  $//img.homeworklib.com/questions/ad842ef0-0fce-11eb-a179-c74271a1379d.png?x-oss-process=image/resize,w_560$ = 31.9375,
GO
= 1.8, n₂ = 16

Z0 =

12 11 of ni 2 2 + 12

=

34.5556-31.9375 3.72 18 + 1.88 16

= 2.6181 / 0.9814

Z0 = 2.6678

P value : 2*P ( z ≥ 2.6678 ) = 2*0.0038

P value = 0.0076

b) 95% confidence interval :

C = 0.95

So α = 1 - C = 1 - 0.95 = 0.05

α/2 = 0.025

Z score correponding to area 0.0250 is 1.96 on the z score table.

Therefore Z = 1.96 ---( from z score table )

Confidence interval is given by,

Lower bound = (
1
- $1596088476556_blob.png$ ) - E  

Upper bound = (
1
- $1596088476669_blob.png$ ) + E  

E = $z*\sqrt{\frac{\sigma _1^2}{n_1}+\frac{\sigma _2^2}{n_2}}$

= $1.96*\sqrt{\frac{3.7^2}{18}+\frac{1.8^2}{16}}$

= 1.9235

(
1
- $1596088476669_blob.png$ ) = 2.6181

Lower bound = (
1
- $1596088476556_blob.png$ ) - E = 2.6181 - 1.9235 = 0.695

Upper bound = (
1
- $1596088476669_blob.png$ ) + E = 2.6181 + 1.9235 = 4.542

0.695  ≤
11
-
112
  ≤ 4.542

c) Given : α = 0.05

As H_a contain ≠  sign , this is two tail test.

α/2 = 0.025

Therefore Z = 1.96

$\sqrt{\frac{\sigma _1^2}{n_1}+\frac{\sigma _2^2}{n_2}}$ = $\sqrt{\frac{3.7^2}{18}+\frac{1.8^2}{16}}$ = 0.9814

Power = P( reject the null hypothesis H0 , when it is false )

$P\left (-1.96\leq \frac{\bar{x}_1 -\bar{x}_2 }{\sqrt{\frac{\sigma _1^2}{n_{1}}+\frac{\sigma _2^2}{n_{2}}}}\leq 1.96 \right )$

= $P\left (-1.96*\sqrt{\frac{\sigma _1^2}{n_{1}}+\frac{\sigma _2^2}{n_{2}}}\leq {\bar{x}_1 -\bar{x}_2 }\leq 1.96*\sqrt{\frac{\sigma _1^2}{n_{1}}+\frac{\sigma _2^2}{n_{2}}} \right )$

= $P\left (-1.96*0.9814\leq {\bar{x}_1 -\bar{x}_2 }\leq 1.96*0.9814 \right )$

= P ( -1.9235  ≤  ${\bar{x}_1 -\bar{x}_2 }$  ≤ 1.9235 )

= $P\left (\frac{-1.9235-2}{0.9814}\leq \frac{(\bar{x}_1 -\bar{x}_2)-(\mu _1-\mu _2) }{\sqrt{\frac{\sigma _1^2}{n_{1}}+\frac{\sigma _2^2}{n_{2}}}}\leq \frac{1.9235 -2}{0.9814}\right )$

= P( -4.0 ≤ z ≤ -0.08 )

= P( z ≤ -0.08 ) - P( z ≤ -4 .0)

= 0.4681 - 0

Power = 0.468