Sondra gets $75 a week, plus 10 percent of sales.

An inspector inspects large truckloads of potatoes to determine the proportion p in the shipment with major defects, prior to us

Sindrei [870]

Answer:

$0.06 - 1.96\sqrt{\frac{0.06(1-0.06)}{200}}=0.027$

$0.06 + 1.96\sqrt{\frac{0.06(1-0.06)}{200}}=0.093$

The 95% confidence interval would be given by (0.027;0.093)

A. 0.027 to 0.093.

Step-by-step explanation:

Notation and definitions

$X=12$ number of defective.

$n=200$ random sample taken

$\hat p=\frac{12}{200}=0.06$ estimated proportion of defectives

$p$ true population proportion of defectives

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

Solution to the problem

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$ . And the critical value would be given by: