Question

Many investors and financial analysts believe the Dow Jones Industrial Average (DJIA) gives a good barometer of the overall stock market. On January 31, 2006, 9 of the 30 stocks making up the DJIA increased in price (The Wall Street Journal, February 1, 2006). On the basis of this fact, a financial analyst claims we can assume that 30% of the stocks traded on the New York Stock Exchange (NYSE) went up the same day. A sample of 69 stocks traded on the NYSE that day showed that 31 went up. You are conducting a study to see if the proportion of stocks that went up is significantly more than 0.3. You use a significance level of α = 0.05 α = 0.05 . What is the test statistic for this sample? (Report answer accurate to three decimal places.) test statistic = What is the p-value for this sample? (Report answer accurate to four decimal places.) p-value = The p-value is... less than (or equal to) α α greater than α α This test statistic leads to a decision to... reject the null accept the null fail to reject the null As such, the final conclusion is that... There is sufficient evidence to warrant rejection of the claim that the proportion of stocks that went up is more than 0.3. There is not sufficient evidence to warrant rejection of the claim that the proportion of stocks that went up is more than 0.3. The sample data support the claim that the proportion of stocks that went up is more than 0.3. There is not sufficient sample evidence to support the claim that the proportion of stocks that went up is more than 0.3.

Answer 1

Testing the hypothesis, it is found that:

• The test statistic is 2.71.
• The p-value is 0.0034.
• The p-value is less than $\alpha$
• The test statistic leads to a decision to reject the null hypothesis.
• The sample data support the claim that the proportion of stocks that went up is more than 0.3.

At the null hypothesis, we test if the proportion of stocks that went up is of 0.3, that is:

$H_0: p = 0.3$

At the alternative hypothesis, we test if the proportion is significantly more than 0.3, that is:

$H_1: p > 0.3$

The test statistic is given by:

$z = \frac{\overline{p} - p}{\sqrt{\frac{p(1-p)}{n}}}$

In which:

• $\overline{p}$ is the sample proportion.
• p is the proportion tested at the null hypothesis.
• n is the sample size.

For this problem, the parameters are:

$p = 0.3, n = 69, \overline{p} = \frac{31}{69} = 0.4493$

The value of the test statistic is:

$z = \frac{\overline{p} - p}{\sqrt{\frac{p(1-p)}{n}}}$

$z = \frac{0.4493 - 0.3}{\sqrt{\frac{0.3(0.7)}{69}}}$

$z = 2.71$

The test statistic is 2.71.

The p-value is the probability of finding a sample proportion of 0.4493 or above, hence, it is 1 subtracted by the p-value of z = 2.71.

Looking at the z-table, z = 2.71 has a p-value of 0.9966.

1 - 0.9966 = 0.0034

The p-value is 0.0034.

Which is less than $\alpha$, and then the test statistic leads to a decision to reject the null hypothesis, and thus:

The sample data support the claim that the proportion of stocks that went up is more than 0.3.

A similar problem is given at https://brainly.com/question/25413788