Stat 211 Handout 9 (Chapter 9): Inferences Based on Two Samples tests

(i) Reject H₀ if z  -z_/2=-1.96 or z  z_/2 =1.96. -1.96< z=1.49 < 1.96 and fail to reject H₀. .

(ii) P-value = 2P(Z>1.49)=2(0.0681)=0.1362. Since the P-value > =0.05, fail to reject H₀.
Conclusion: Yes it is reasonable to assume that both machines produce the same fraction of defective parts
The MINITAB output analyzing such data is

Test and CI for Two Proportions

Sample X N Sample p

1 15 300 0.050000

2 8 300 0.026667

Estimate for p(1) - p(2): 0.0233333

95% CI for p(1) - p(2): (-0.00733568, 0.0540023)

Test for p(1) - p(2) = 0 (vs not = 0): Z = 1.49 P-Value = 0.136

Test of p = 0.04 vs p not = 0.04

Sample X N Sample p 95.0% CI Exact P-Value

1 15 300 0.050000 (0.028251, 0.081127) 0.461

2 8 300 0.026667 (0.011582, 0.051866) 0.247
Example 6 (Exercise 9.48(a)):

Sample sizes are large enough to satisfy the assumptions and it is a two-tailed test.

Decision:

(i) Reject H₀ if z  -z_/2=-1.96 or z  z_/2 =1.96. |z| =4.844 > 1.96 and reject H₀.

(ii) P-value = 2P(Z>4.844)=2(0)=0. Since the P-value  =0.05, reject H₀.

Test and CI for Two Proportions

Sample X N Sample p

1 63 300 0.210000

2 75 180 0.416667

Estimate for p(1) - p(2): -0.206667

95% CI for p(1) - p(2): (-0.292174, -0.121159)

Test for p(1) - p(2) = 0 (vs not = 0): Z = -4.74 P-Value = 0.000

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