Question

The Ohio Department of Agriculture tested 203 fuel samples across the state

Answer 1

Answer:

99% confidence level for the true population proportion of premium grade fuel-quality failures is between a lower limit of 0.048 and an upper limit of 0.218.

Step-by-step explanation:

confidence interval for a proportion is given as p +/- margin of error (E)

p is sample proportion = 14/105 = 0.133

n is number of samples = 105

confidence level (C) = 99% = 0.99

significance level = 1 - C = 1 - 0.99 = 0.01 = 1%

At 1% significance level, critical value (z) is 2.576

Margin of error (E) = zsqrt [p(1-p) ÷ n] = 2.576sqrt[0.133(1-0.133) ÷ 105] = 2.576 × 0.0331391 = 0.085

Lower limit of proportion = p - E = 0.133 - 0.085 = 0.048

Upper limit of proportion = p + E = 0.133 + 0.085 = 0.218

99% confidence level is (0.048, 0.218)

Answer 2

Answer:

$\hat p = \frac{14}{105}= 0.133$

And that represent the proportion of failures.

The confidence interval for the mean is given by the following formula:  

$\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$

If we replace the values obtained we got:

$0.133 - 2.58\sqrt{\frac{0.133(1-0.133)}{105}}=0.0475$

$0.133 + 2.58\sqrt{\frac{0.133(1-0.133)}{105}}=0.2185$

The 99% confidence interval would be given by (0.0475;0.2185)

Step-by-step explanation:

Previous concept

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 99% of confidence, our significance level would be given by $\alpha=1-0.99=0.01$ and $\alpha/2 =0.005$. And the critical value would be given by:

$z_{\alpha/2}=-2.58, z_{1-\alpha/2}=2.58$

The proportion estimated would be:

$\hat p = \frac{14}{105}= 0.133$

And that represent the proportion of failures.

The confidence interval for the mean is given by the following formula:  

$\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$

If we replace the values obtained we got:

$0.133 - 2.58\sqrt{\frac{0.133(1-0.133)}{105}}=0.0475$

$0.133 + 2.58\sqrt{\frac{0.133(1-0.133)}{105}}=0.2185$

The 99% confidence interval would be given by (0.0475;0.2185)