In: Statistics and Probability
ACE appliance manufacturing company claims that the proportion of defectives in their brand is lower than that of PREMIER’s, their next door competitor. The manager of ACE wants to demonstrate the validity of this claim. For this purpose, he selects a random sample of 36 appliances from each company and find that 3 ACE and 5 PREMIER’s appliances are defective. Should we believe ACE’s claim based on this data?
(a) Let ACE be population 1 and PREMIER be population 2. State the null hypothesis and alternative hypothesis
(b) Find the test statistic
(c) If the type-I error α = 0.05, find the critical value(s) and shade the rejection region(s)
(d) Base on the type-I error α and rejection region(s), given above, what is your conclusion?
For sample 1, we have that the sample size is N_1= 36 , the number of favorable cases is X_1 = 3 , so then the sample proportion is
For sample 2, we have that the sample size is N_2 = 36, the number of favorable cases is X_2 = 5 , so then the sample proportion is
The value of the pooled proportion is computed as
Also, the given significance level is α=0.05.
(1) Null and Alternative Hypotheses
The following null and alternative hypotheses need to be tested:
Ho: p_1 = p_2
Ha: p_1 < p_2
This corresponds to a left-tailed test, for which a z-test for two population proportions needs to be conducted.
(2) Rejection Region
Based on the information provided, the significance level is α=0.05, and the critical value for a left-tailed test is z_c = -1.64
(3) Test Statistics
The z-statistic is computed as follows:
(4) Decision about the null hypothesis
Since it is observed that z = -0.75 ≥zc=−1.64, it is then concluded that the null hypothesis is not rejected.
Using the P-value approach: The p-value is p = 0.2266, and since p = 0.2266 ≥0.05, it is concluded that the null hypothesis is not rejected.
(5) Conclusion
It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population proportion p1 is less than p2, at the 0.05 significance level.