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3 years ago

Umm please help me wit this problem.

Mathematics

1 answer:

Afina-wow [57]3 years ago

3 0

Answer:

1. 11

2. 3

Step-by-step explanation:

Hope this helps!! Have An Amazing Day!! Let me know if this is wrong.

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A recent article in a university newspaper claimed that the proportion of students who commute more than miles to school is no m

Naddika [18.5K]

Answer:

The null hypothesis is $H_0: p \leq x$ , in which x is the proportion tested.

The alternative hypothesis is $H_1: p > x$

Step-by-step explanation:

A recent article in a university newspaper claimed that the proportion of students who commute more than miles to school is no more than x.

This means that at the null hypothesis, we test if the proportion is of at most x, that is:

$H_0: p \leq x$

Suppose that we suspect otherwise and carry out a hypothesis test.

The opposite of at most x is more than x, so the alternative hypothesis is:

$H_1: p > x$

4 0

3 years ago

A banana has 80 calories. This is 5 calories less than one seventh of the calories in a banana split. How many calories in a ban

viva [34]

The answer is 85 times 7. Since the banana was 5 less then the split, the split is 5 more which is 85 calories but then it says it is also 1/7 of the whole split. So you simply multiply 85 and 7 because 85 was only 1/7 of the total banana split

4 0

3 years ago

Read 2 more answers

The amounts (in ounces) of randomly selected eight 16-ounce beverage cans are given below. See Attached Excel for Data. Assume t

motikmotik

The question is incomplete! Complete question along with answer and step by step explanation is provided below.

Question:

The amounts (in ounces) of randomly selected eight 16-ounce beverage cans are given below.

16.5, 15.2, 15.4, 15.1, 15.3, 15.4, 16, 15.1

Assume that the amount of beverage in a randomly selected 16-ounce beverage can has a normal distribution. Compute a 99% confidence interval for the population mean amount of beverage in 16-ounce beverage cans and fill in the blanks appropriately.

A 99% confidence interval for the population mean amount of beverage in 16-ounce beverage cans is ( , ) ounces. (round to 3 decimal places)

Answer:

$99\% \: \text {confidence interval} = (14.886, \: 16.113)\\\\$

Therefore, the 99% confidence interval for the population mean amount of beverage in 16-ounce beverage cans is (14.886, 16.113) ounces.

Step-by-step explanation:

Let us find out the mean amount of the 16-ounce beverage cans from the given data.

Using Excel,

=AVERAGE(number1, number2,....)

The mean is found to be

$\bar{x} = 15.5$

Let us find out the standard deviation of the 16-ounce beverage cans from the given data.

Using Excel,

=STDEV(number1, number2,....)

The standard deviation is found to be

$s = 0.4957$

The confidence interval is given by

$\text {confidence interval} = \bar{x} \pm MoE\\\\$

Where $\bar{x}$ is the sample mean and Margin of error is given by

$$ MoE = t_{\alpha/2} \cdot (\frac{s}{\sqrt{n} } ) $ \\\\$

Where n is the sample size, s is the sample standard deviation and is the t-score corresponding to a 99% confidence level.

The t-score corresponding to a 99% confidence level is

Significance level = α = 1 - 0.99 = 0.01/2 = 0.005

Degree of freedom = n - 1 = 8 - 1 = 7

From the t-table at α = 0.005 and DoF = 7

t-score = 3.4994

$MoE = t_{\alpha/2}\cdot (\frac{s}{\sqrt{n} } ) \\\\MoE = 3.4994 \cdot \frac{0.4957}{\sqrt{8} } \\\\MoE = 3.4994\cdot 0.1753\\\\MoE = 0.6134\\\\$

So the required 99% confidence interval is

$\text {confidence interval} = \bar{x} \pm MoE\\\\\text {confidence interval} = 15.5 \pm 0.6134\\\\\text {confidence interval} = 15.5 - 0.6134, \: 15.5 + 0.6134\\\\\text {confidence interval} = (14.886, \: 16.113)\\\\$