Question

A sample of 44 observations is selected from one population with a population standard deviation of 3.1. The sample mean is 101.0. A sample of 56 observations is selected from a second population with a population standard deviation of 5.0. The sample mean is 99.5. Conduct the following test of hypothesis using the 0.10 significance level.

H₀ : μ₁ = μ₂

H₁ : μ₁ ≠ μ₂

1. Is this a one-tailed or a two-tailed test?

• One-tailed test

• Two-tailed test

2. State the decision rule. (Negative values should be indicated by a minus sign. Round your answers to 2 decimal places.)

3. Compute the value of the test statistic. (Round your answer to 2 decimal places.)

4. What is your decision regarding H₀?

• Reject H₀

• Do not reject H₀

5. What is the p-value? (Round your answer to 4 decimal places.)

Answer 1

n,=44 x= 101.0 0,=31 M2=56 X2=99.5 = 50 POPULATION STANDARD DEVIATION Xi = SAMPLE MEAN OF SAMI n - SAMPLE SIZES NULL HYPOTHESIS Mo:M, = M₂ ALTERNATE HYPOTHESIS H,: M, M2 (A) SINCE IN ABOVE HYPOTHESIS, ONLY A DIFFERENCE S EXPECTED THE INVESTIGATOR DOES NOT HXROTH MAKE PRETO comMENTS ON WHETHER POPULATION MEAN IS GREATER OR LESSER THAN POPOLATION MEAN FOR THE OTHER. THEREFORE, IT IS A Two TALLER TEST IT WOULD HAVE BEEN ONE TAILED TEST OF ALTERNATE HYPOTHESIS o WOULD HAVE BEEN - H: M, Z M2 Hi: M, < u2 chalen SAMPLES FROM POPULATIONS OF SIZE n a nz with population means (unknown M, & M2 and known standard deviations G & O2, CS Scates twith statistic comparing the means CamScanner

known as two - sample 2- statistic 2=(5,-26? -(4,-M2) No.2 + 02² ni nz NORMAL which has standard 1. distribution (N,(0,1)) CASE SINCE ABOVE HYPOTH IN ABOVE 2 - tailed test is proposed PECSON RULE is: REJECT Ho CNULL HYPOTHESIS) IF 25-1.960 OR IF 2 Z 1.960 CSMALLER THAN LARGER THAN LOWER T of UPPER CRITICAL VALUE CRITICAL VALUE) HERE x=0-os (ASSUMED c) test stastic 2=(x - ez? -(", -42) ni Scanned with CamScanner

(101.o 2- Eteo -99.5) - (0) (3.12 (5.02 44 s6 2 = 15 = los d 0.2184 +0.446 No. 66482 6.81536 2= 1.8396 = 1.83 (9) WE HAVE TO TEST ABOVEE MY POTHESIS USING 010 SIGNIFICANCE LEVEL SINCE ABOVE TWO-TAIL TEST IS PROPOSED UPPER CRITICAL VALUE = 2012 = 1,645 LOWER CRITICAL VALUE = 2-1/2 = -10645 SINCE IT is Two TALL Tra=0.90 FOR CRITICAL Now let WE TEEP CHECK VALUES IN 2- TABLE at (1-2) = 0.45 Scanned with CamScanner

> 20.os 2o-os ..21183 2-4/2 1.645 -1.645 REJECTION REGION FORHO ACCEPTANCE REGION FOR Ho REJECTION REGION FOR Ho ce pevalue THEREFORE, WEE REPECT Ho CNOLL HYPOTHESIS THAT POPULATION POPULATIONS MEANS ARE EQUAL el p-value for above z-statistic le 2=1.83 is C1-0.9664 = 0,03367 2=1. 83 Scanned with CamScanner

Answer 2

Given that n = 44 J = 31 Ž, - 101.0 ₂ = 56 102=5.0 1x, = 99.5 | 2:0.10 Answer of Two tailed test (6) At 2 = 0.10 Lois (tworlailed) Z-C.V = & 1.640 decision sule : reject to if Z<[1.64 (08) Z311.64 (c) Test statistic: Z= -* - 101.0 - 99.5 uno 0.815376 = 1.84 z = 1.84 (d) Tead> Zevine, 1.843 1.64 so reject Ho] ce) with Z=1.84 (two tailed). we get P-value = 0.0658]