Question

2. When a large truckload of mangoes arrives at a packing plant, a random sample of 150 is selected and examined for

Answer 1

a) The 90% confidence interval of the percentage of all mangoes on the truck that fail to meet the standards is: (7.55%, 12.45%).

b) The margin of error is: 2.45%.

c) The 90% confidence is the level of confidence that the true population percentage is in the interval.

d) The needed sample size is: 271.

What is a confidence interval of proportions?

A confidence interval of proportions has the bounds given by the rule presented as follows:

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

The margin of error is given by:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

The variables are listed as follows:

• $\pi$ is the sample proportion, which is also the estimate of the parameter.

• z is the critical value.

• n is the sample size.

The confidence level is of 90%, hence the critical value z is the value of Z that has a p-value of $\frac{1+0.90}{2} = 0.95$, so the critical value is z = 1.645.

The sample size and the estimate are given as follows:

$n = 150, \pi = \frac{15}{150} = 0.1$

The margin of error is of:

$M = z\sqrt{\frac{0.1(0.9)}{150}} = 0.0245 = 2.45\%$

The interval is given by the estimate plus/minus the margin of error, hence:

• The lower bound is: 10 - 2.45 = 7.55%.

• The upper bound is: 10 + 2.45 = 12.45%.

For a margin of error of 3% = 0.03, the needed sample size is obtained as follows:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.03 = 1.645\sqrt{\frac{0.1(0.9)}{n}}$

$0.03\sqrt{n} = 1.645\sqrt{0.1(0.9)}$

$\sqrt{n} = \frac{1.645\sqrt{0.1(0.9)}}{0.03}$

$(\sqrt{n}})^2 = \left(\frac{1.645\sqrt{0.1(0.9)}}{0.03}\right)^2$

n = 271 (rounded up).

More can be learned about the z-distribution at brainly.com/question/25890103

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