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mezya [45]
3 years ago
14

Someone please help me this is the last question on my homework. I have to write a quadratic equation for this word problem...

Mathematics
1 answer:
antiseptic1488 [7]3 years ago
5 0

Answer:

57xsqaure-42x=5.5

Step-by-step explanation:

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A university financial aid office polled a random sample of 670 male undergraduate students and 617 female undergraduate student
ahrayia [7]

Answer:

The 90% confidence interval for the difference between the proportions of male and female students who were employed during the summer is (0.01, 0.1012).

Step-by-step explanation:

Before building the confidence interval, we need to understand the central limit theorem and subtraction of normal variables.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean \mu = p and standard deviation s = \sqrt{\frac{p(1-p)}{n}}

Subtraction between normal variables:

When two normal variables are subtracted, the mean is the difference of the means, while the standard deviation is the square root of the sum of the variances.

Male undergraduates:

670, 388 were employed. So

p_M = \frac{388}{670} = 0.5791

s_M = \sqrt{\frac{0.5791*0.4209}{670}} = 0.0191

Female undergraduates:

Of 617, 323 were employed. So

p_F = \frac{323}{617} = 0.5235

s_F = \sqrt{\frac{0.5235*0.4765}{617}} = 0.0201

Distribution of the difference:

p = p_M - p_F = 0.5791 - 0.5235 = 0.0556

s = sqrt{s_M^2+s_F^2} = \sqrt{0.0201^2 + 0.0191^2} = 0.0277

Confidence interval:

The confidence interval is given by:

p \pm zs

In which

z is the z-score that has a p-value of 1 - \frac{\alpha}{2}.

90% confidence level

So \alpha = 0.1, z is the value of Z that has a p-value of 1 - \frac{0.1}{2} = 0.95, so Z = 1.645.

The lower bound of the interval is:

p - zs = 0.0556 - 1.645*0.0277 = 0.01

The upper bound of the interval is:

p + zs = 0.0556 + 1.645*0.0277 = 0.1012

The 90% confidence interval for the difference between the proportions of male and female students who were employed during the summer is (0.01, 0.1012).

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3 years ago
A _ is a ratio that compares a number to 100 using the % symbol
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A percentage is a ratio that compares a number to 100 using the % symbol
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