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3 years ago

In a set of ordered pairs, the y values repeated while the x values did not repeat.

Mathematics

1 answer:

LiRa [457]3 years ago

7 0

Answer: D

Step by Step: Even though the y value repeats, to be considered not a function is when the X value repeats

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2 years ago

For a large supermarket chain in a particularâ state, aâ women's group claimed that female employees were passed over for manage

PolarNik [594]

Answer:

1) As the P-value (0.097) is greater than the significance level (0.05), the effect is not significant.

The null hypothesis failed to be rejected.

There is not enough evidence to support the women's claim that female employees were passed over for management training in favor of their male colleagues.

2) The 95% confidence interval for the population proportion is (0.173, 0.427).

The populations proportion (π=0.4) is included in this confidence interval, so the claim has no enough evidence to be supported.

Step-by-step explanation:

This is a hypothesis test for a proportion.

The claim is that female employees were passed over for management training in favor of their male colleagues.

We use the women's part for sample and population proportions.

Then, the null and alternative hypothesis are:

$H_0: \pi=0.4\\\\H_a:\pi$

The significance level is 0.05.

The sample has a size n=50 persons.

The sample proportion is p=0.3.

$p=X/n=15/50=0.3$

The standard error of the proportion is:

$\sigma_p=\sqrt{\dfrac{\pi(1-\pi)}{n}}=\sqrt{\dfrac{0.4*0.6}{50}}\\\\\\ \sigma_p=\sqrt{0.0048}=0.069$

Then, we can calculate the z-statistic as:

$z=\dfrac{p-\pi+0.5/n}{\sigma_p}=\dfrac{0.3-0.4+0.5/50}{0.069}=\dfrac{-0.09}{0.069}=-1.299$

This test is a left-tailed test, so the P-value for this test is calculated as:

$\text{P-value}=P(z$

As the P-value (0.097) is greater than the significance level (0.05), the effect is not significant.

The null hypothesis failed to be rejected.

There is not enough evidence to support the claim that female employees were passed over for management training in favor of their male colleagues.

b) We have to calculate a 95% confidence interval for the proportion.

The sample proportion is p=0.3.

The standard error of the proportion is:

$\sigma_p=\sqrt{\dfrac{p(1-p)}{n}}=\sqrt{\dfrac{0.3*0.7}{50}}\\\\\\ \sigma_p=\sqrt{0.0042}=0.0648$

The critical z-value for a 95% confidence interval is z=1.96.

The margin of error (MOE) can be calculated as:

$MOE=z\cdot \sigma_p=1.96 \cdot 0.0648=0.127$

Then, the lower and upper bounds of the confidence interval are:

$LL=p-z \cdot \sigma_p = 0.3-0.127=0.173\\\\UL=p+z \cdot \sigma_p = 0.3+0.127=0.427$

The 95% confidence interval for the population proportion is (0.173, 0.427).

The populations proportion (π=0.4) is included in this confidence interval, so the claim has no enough evidence to be supported.

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3 years ago