Question

I'll give brainiest, 1. Out of a sample of 150 people who walked into a fast-food restaurant, 135 said they prefer ketchup to mustard on their hamburgers. With 89% confidence, what is the confidence interval for the population mean of customers that prefer ketchup to mustard on their hamburgers?

CI = (86.09%, 93.91%)
CI = (85.20%, 94.80%)
CI = (82.61%, 95.87%)
CI = (83.69%, 96.31)

2. Out of all the students that attend Yale University, 1,982 were selected to participate in a school survey about their commute to school. 70% said that their commute to school was fine, while the others were dissatisfied with their commute. Calculate a 90% confidence interval for the proportion of students at Yale who feel their commute to school is fine the way it is.

CI = (68.76%, 71.98%)
CI = (67.35%, 72.65%)
CI = (68.31%, 71.69%)
CI = (67.98%, 72.02%)

3. Out of a sample of 500 people, 25% exercise daily. What is the confidence interval for the population mean of people that exercise daily, given a confidence level of 95%?

CI = (20.98%, 30.02%)
CI = (21.81%, 28.19%)
CI = (21.20%, 28.80%)
CI = (20.01%, 29.99%)

4. The height of pyramids in Egypt is normally distributed. A sample of 125 pyramids had a mean height of 460.7 ft and a standard deviation of 4.1 feet. With 80% confidence, what is the maximum error of estimate for the actual population mean for the height of Egyptian pyramids?

37.04%
47.01%
4.21%
94.43%

Answer 1

The confidence intervals are given by:

1. CI = (86.09%, 93.91%)

2. CI = (68.31%, 71.69%)

3. CI = (21.20%, 28.80%)

In 4.,  the margin of error is of 47.01%.

What is a confidence interval of proportions?

A confidence interval of proportions is given by:

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

In which:

• $\pi$ is the sample proportion.

• z is the critical value.

• n is the sample size.

Item 1:

The parameters are:

$z = 1.5982, n = 150, \pi = \frac{135}{150} = 0.9$

The lower bound of the interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.9 - 1.5982\sqrt{\frac{0.9(0.1)}{150}} = 0.8609$

The upper bound of the interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.9 + 1.5982\sqrt{\frac{0.9(0.1)}{150}} = 0.9391$

Item 2:

The parameters are:

$z = 1.645, n = 1982, \pi = 0.7$

The lower bound of the interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.7 - 1.645\sqrt{\frac{0.7(0.3)}{1982}} = 0.6831$

The upper bound of the interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.7 + 1.645\sqrt{\frac{0.7(0.3)}{1982}} = 0.7169$

Item 3:

The parameters are:

$z = 1.96, n = 500, \pi = 0.25$

The lower bound of the interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.25 - 1.96\sqrt{\frac{0.25(0.75)}{500}} = 0.2120$

The upper bound of the interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.25 + 1.96\sqrt{\frac{0.25(0.75)}{500}} = 0.2880$

Item 4:

We have the standard deviation for the sample, hence the t-distribution is used, and the margin of error is given by:

$M = t\frac{s}{\sqrt{n}}$

The parameters are:

$t = 1.2884, s = 4.1, n = 125$

Hence:

$M = t\frac{s}{\sqrt{n}}$

$M = 1.2884\frac{4.1}{\sqrt{125}}$

$M = 0.4701$

47.01% is the margin of error.

More can be learned about confidence intervals at https://brainly.com/question/16162795