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2 years ago

-v+5+6v=1+5v+3 solve for v

Mathematics

1 answer:

earnstyle [38]2 years ago

3 0

Answer: No solution for V. No solution.

Step-by-step explanation:

-v+5+6v=1+5v+3

Combine like terms together by adding all of the v's together, and number together.

5v+5=4+5v

5v-5v+5=4

0+5=4

5=4

Since this is a false statement, then no solution for V

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1. There is not enough evidence to support the claim that bullish sentiment differs from its long-term average of 0.39.

2. There is enough evidence to support the claim that bearish sentiment is above its long-term average of 0.30.

Step-by-step explanation:

The question is incomplete:

The American Association of Individual Investors conducts a weekly survey of its members to measure the percent who are bullish, bearish, and neutral on the stock market for the next six months. For the week ending November 7, 2012 the survey results showed 38.5% bullish, 21.6% neutral, and 39.9% bearish (AAII website, November 12, 2012). Assume these results are based on a sample of 300 AAII members.

1. This is a hypothesis test for a proportion.

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$H_0: \pi=0.39\\\\H_a:\pi\neq 0.39$

The significance level is 0.05.

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The sample proportion is p=0.385.

The standard error of the proportion is:

$\sigma_p=\sqrt{\dfrac{\pi(1-\pi)}{n}}=\sqrt{\dfrac{0.39*0.61}{300}}\\\\\\ \sigma_p=\sqrt{0.000793}=0.028$

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

$P-value=2\cdot P(z$

As the P-value (0.906) 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 bullish sentiment differs from its long-term average of 0.39.

2) This is a hypothesis test for a proportion.

The claim is that bearish sentiment is above its long-term average of 0.30.

Then, the null and alternative hypothesis are:

$H_0: \pi=0.3\\\\H_a:\pi\neq 0.3$

The significance level is 0.05.

The sample has a size n=300.

The sample proportion is p=0.399.

The standard error of the proportion is:

$\sigma_p=\sqrt{\dfrac{\pi(1-\pi)}{n}}=\sqrt{\dfrac{0.3*0.7}{300}}\\\\\\ \sigma_p=\sqrt{0.0007}=0.026$

Then, we can calculate the z-statistic as:

$z=\dfrac{p-\pi-0.5/n}{\sigma_p}=\dfrac{0.399-0.3-0.5/300}{0.026}=\dfrac{0.097}{0.026}=3.679$

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

$P-value=2\cdot P(z>3.679)=0$

As the P-value (0) is smaller than the significance level (0.05), the effect is significant.

The null hypothesis is rejected.

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