Question

1. A new rocket-launching system is being considered for deployment of small, short-range rockets. A sample of 80 experimental launches is made with the new system, and 54 are successful. (a) Create a 98% confidence interval for the proportion of successful launches for the new rocketlaunching system. (b) The existing system has p

Answer 1

Answer:

The 98% confidence interval for the proportion of successful launches for the new rocket launching system is (0.553, 0.797).

Step-by-step explanation:

We have to calculate a 98% confidence interval for the proportion.

The sample proportion is p=0.675.

$p=X/n=54/80=0.675$

The standard error of the proportion is:

$\sigma_p=\sqrt{\dfrac{p(1-p)}{n}}=\sqrt{\dfrac{0.675*0.325}{80}}\\\\\\ \sigma_p=\sqrt{0.00274}=0.052$

The critical z-value for a 98% confidence interval is z=2.326.

The margin of error (MOE) can be calculated as:

$MOE=z\cdot \sigma_p=2.326 \cdot 0.052=0.122$

Then, the lower and upper bounds of the confidence interval are:

$LL=p-z \cdot \sigma_p = 0.675-0.122=0.553\\\\UL=p+z \cdot \sigma_p = 0.675+0.122=0.797$

The 98% confidence interval for the population proportion is (0.553, 0.797).