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kodGreya [7K]
3 years ago
7

Jason owns two square plot of land that have the same dimensions. He subdivides one into 10 equal parcels and the other into 11

equal parcels. For the same price you can buy either 3 of the 10 parcels or 4 of the 11 parcels. Do the two parcels yield the same amount of land? Can anyone show me how to do this step by step?
Mathematics
1 answer:
Ymorist [56]3 years ago
4 0
No, they do not yield the same amount of land.

If you divide a number by 10, you get a larger result than if you divide it by 11. So let's say the area of the two squares is 110:

110/10 = 11 for each parcel

110/11 = 10 for each parcel
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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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Step-by-step explanation:

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Answer:

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