Respuesta :
Answer:
The 95% confidence interval for the true proportion of defective batteries is (0.0966, 0.1034).
It is better to take a larger sample to derive conclusion about the true parameter value.
Step-by-step explanation:
The (1 - α) % confidence interval for proportion is:
[tex]CI=\hat p\pm z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex]
Given:
n = 2000
X = 200
The sample proportion is:
[tex]\hat p=\frac{X}{n}=\frac{200}{2000}=0.10[/tex]
The critical value of z for 95% confidence interval is:
[tex]z_{\alpha /2}=z_{0.05/2}=z_{0.025}=1.96[/tex]
Compute the 95% confidence interval as follows:
[tex]CI=\hat p\pm z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}\\=0.10\pm1.96\times\sqrt{\frac{0.10(1-0.10)}{2000}}\\=0.10\pm0.0034\\=(0.0966, 0.1034)[/tex]
Thus, the 95% confidence interval for the true proportion of defective batteries is (0.0966, 0.1034).
Now if in a sample of 100 batteries there are 15 defectives, the the 95% confidence interval for this sample is:
[tex]CI=\hat p\pm z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}\\=0.15\pm1.96\times\sqrt{\frac{0.15(1-0.15)}{100}}\\=0.15\pm0.0706\\=(0.0794, 0.2206)[/tex]
It can be observed that as the sample size was decreased the width of the confidence interval was increased.
Thus, it can be concluded that it is better to take a larger sample to derive conclusion about the true parameter value.
