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 bullish, neutral, and bearish (AAII website, November 12, 2012). Assume these results are based on a sample of AAII members. a. Over the long-term, the proportion of bullish AAII members is . Conduct a hypothesis test at the level of significance to see if the current sample results show that bullish sentiment differs from its long-term average of . What are your findings

p-value = 0.8572

The current sample results do not show that bullish sentiment differs from its long-term average of 0.39, that is, the sample results show that there is no difference between the current bullish sentiments and the long-term average of 0.39.

Step-by-step explanation:

For hypothesis testing, the first thing to define is the null and alternative hypothesis.

The null hypothesis plays the devil's advocate and usually takes the form of the opposite of the theory to be tested. It usually contains the signs =, ≤ and ≥ depending on the directions of the test.

While, the alternative hypothesis usually confirms the the theory being tested by the experimental setup. It usually contains the signs ≠, < and > depending on the directions of the test.

For this question, we want to test if the current sample results show that bullish sentiment differs from its long-term average of 0.39.

The null hypothesis would be that there isn't enough evidence from the sample results to suggest that the bullish sentiment differs from its long-term average of 0.39.

The alternative hypothesis is that there is enough evidence from the sample results to suggest that the bullish sentiment differs from its long-term average of 0.39.

Mathematically, if the population proportion of bullish members for that period/week is p.

The null hypothesis is represented as

H₀: p = 0.39

The alternative hypothesis is represented as

Hₐ: p ≠ 0.39

To do this test, we will use the z-distribution because although, no information on the population standard deviation is known, 5he sample size is large enough for the sample properties to approximate the population properties.

So, we compute the z-test statistic

z = (x - μ)/σₓ

x = sample proportion of bullish members = 0.385

μ = p₀ = the standard we are comparing against = 0.39

σₓ = standard error = √[p(1-p)/n]

where n = Sample size = 300

σₓ = √[0.385×0.615/300] = 0.0280935936 = 0.02809

z = (0.385 - 0.39) ÷ 0.02809

z = -0.1779765192 = -0.18

checking the tables for the p-value of this z-statistic

Significance level = 0.05

The hypothesis test uses a two-tailed condition because we're testing in both directions.

p-value (for t = 0.18, at 0.05 significance level, with a two tailed condition) = 0.857153

The interpretation of p-values is that

When the (p-value > significance level), we fail to reject the null hypothesis and when the (p-value < significance level), we reject the null hypothesis and accept the alternative hypothesis.

So, for this question, significance level = 0.05

p-value = 0.857153

0.857153 > 0.05

Hence,

p-value > significance level

This means that we fail to reject the null hypothesis & say that there isn't enough evidence from the sample results to suggest that the bullish sentiment differs from its long-term average of 0.39.

Hope this Helps!!!

529 and 531

step-by-step explanation:

g(x) = 3*(8(3^(x−2))+2)

g(x) = 24 (3^(x-2) )+6

the factor is 2

step-by-step explanation:

let f(x)=8(3^(x−2)+2)

to stretch a graph by 3, it would be 3f(x)

g(x) =3f(x)

g(x) = 3*(8(3^(x−2))+2)

g(x) = 24 (3^(x-2) )+6

x     −2     0     2     4     6

f(x) 1/64 1/16   1/4     1     4

1/64 * z * z = 1/16

i multiplies by z twice because it its 2 steps from -2 to 0

1/64 z^2 = 1/16

multiply by 64 to clear the fraction on the left side

1/4 * 64 z^2 =1/16 *64

z^2 = 4

take the square root of each side

sqrt(z^2) = ±sqrt(4)

z = ±2  

we are multiplying by 2 each time.