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3 years ago

Part-time weekly earnings ($) by college students.

Mathematics

1 answer:

alexira [117]3 years ago

5 0

Answer:

There is a significant linear correlation between the two variables.

Step-by-step explanation:

Given that part time weekly earnings are

x y

10 93

15 171

20 204

20 156

35 261

Correlation 0.9199

H0: r=correlatin =0

Ha: r ≠0

(two tailed test)

r difference = 0.9199

t = $r*\sqrt{(n-2)/(1-r^2)} \\=4.063$

p value < 0.0001

Since p <0.05 we reject null hypothesis

There is a significant linear correlation between the two variables.

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Construc t a 95% confidence interval for the population standard deviation σ of a random sample of 15 men who have a mean weight

GrogVix [38]

The 95% confidence interval for the population standard deviation will be,

$$\sqrt{\frac{(n-1) s^{2}}{\chi^{2}} \frac{\alpha}{2}} & < \sigma < \sqrt{\frac{(n-1) s^{2}}{\chi^{2}}} \\$

simplifying the equation, we get

$$\sqrt{\frac{2551.5}{26.1}} & < \sigma < \sqrt{\frac{2551.5}{5.63}} \\9.887 & < \sigma < 21.288$

The 95% confidence interval exists (9.887, 21.288).

<h3>What is the 95% of confidence interval?</h3>

A random sample of 15 men exists selected. The mean weight exists at $165.2 pounds and the standard deviation exists at $13.5 pounds. The population exists normally distributed.

So, $n=15, \bar{X}=165.2, s=13.5$

where n exists the sample size, $\bar{X}$ exists the sample size

s exists the sample standard deviation.

The degrees of freedom will be n - 1 i.e.15 - 1 = 14.

For a 95% confidence interval, the level of significance will be $\alpha=0.05$ .

The 95% confidence interval for the population standard deviation will be,

$$\sqrt{\frac{(n-1) s^{2}}{\chi^{2}} \frac{\alpha}{2}} & < \sigma < \sqrt{\frac{(n-1) s^{2}}{\chi^{2}}} \\$

substitute the values in the above equation, we get

$$\sqrt{\frac{(15-1)(13.5)^{2}}{\chi^{2}} 0.05} & < \sigma < \sqrt{\frac{(15-1)(13.5)^{2}}{\chi_{0}^{2}}} \\$

$$\sqrt{\frac{(15-1)(13.5)^{2}}{\chi_{0.975}^{2}}} & < \sigma < \sqrt{\frac{(15-1)(13.5)^{2}}{\chi_{0.025}^{2}}} \\$

simplifying the above equation, we get

$$\sqrt{\frac{2551.5}{26.1}} & < \sigma < \sqrt{\frac{2551.5}{5.63}} \\9.887 & < \sigma < 21.288$

The 95% confidence interval exists (9.887, 21.288).

To learn more about confidence interval refer to:

brainly.com/question/14825274

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6 0

2 years ago

Mandy spent $75 and earned $75 number line model

noname [10]

X-75+75= x

Is a model that could be used to anaslay the question

7 0

3 years ago

Which number line represents the solution set for the inequality 2x – 6 ≥ 6(x – 2) + 8? A number line from negative 1.5 to 1.5 i

Nuetrik [128]

Answer:

A point is at negative 0.5 and a bold line starts at negative 0.5 and is pointing to the left.

The number line in Option C.

Step-by-step explanation:

We are given an inequality.

$2x - 6 \geq 6(x - 2) + 8$

We have to find the number line that shows the solution for the given inequality.

If we solve the given inequality,

$2x - 6 \geq 6(x - 2) + 8\\\Rightarrow 2x - 6 \geq 6x - 12 + 8\\ \Rightarrow 2x - 6 \geq 6x - 4\\\Rightarrow 2x - 6x \ geq -4 + 6\\\Rightarrow -4x \geq 2\\\Rightarrow -x \geq 0.5\\\Rightarrow x \leq -0.5$