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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 74 stocks traded on the NYSE that day showed that 28 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 $\alpha=0.05$.

What is the test statistic for this sample? (Report answer accurate to three decimal places.) test statistic $=1.471 \checkmark 0^{\circ}$

What is the p-value for this sample? (Report answer accurate to four decimal places.)
\[
p \text {-value }=0.3784 x
\]

The p-value is...
less than (or equal to) $\alpha$

Step 1 ：The problem is asking for the test statistic and p-value for a sample of 74 stocks, where 28 went up. This is a hypothesis testing problem for a proportion. The null hypothesis is that the proportion of stocks that went up is 0.3, and the alternative hypothesis is that the proportion is more than 0.3.

Step 2 ：To calculate the test statistic, we need to use the formula for the z-score in hypothesis testing for a proportion, which is \((p_{hat} - p_0) / \sqrt{(p_0 * (1 - p_0)) / n}\), where \(p_{hat}\) is the sample proportion, \(p_0\) is the hypothesized population proportion, and n is the sample size.

Step 3 ：Given that n = 74, x = 28, \(p_0\) = 0.3, and \(\alpha\) = 0.05, we can calculate \(p_{hat}\) as 0.3783783783783784.

Step 4 ：Substituting these values into the formula, we get a z-score of approximately 1.471.

Step 5 ：To calculate the p-value, we need to find the probability of getting a z-score as extreme as the one we calculated, under the null hypothesis. This is done by finding the area under the standard normal curve to the right of the calculated z-score.

Step 6 ：The calculated p-value is approximately 0.071.

Step 7 ：The p-value is greater than the significance level of 0.05, which means we do not reject the null hypothesis that the proportion of stocks that went up is 0.3.

Step 8 ：Final Answer: The test statistic for this sample is approximately \(\boxed{1.471}\). The p-value for this sample is approximately \(\boxed{0.071}\).