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Many investors and financial analysts believe the Dow Jones Industrial Average (DJIA) provides 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. Formulate null and alternative hypotheses to test the analyst's claim. b. \(\quad\) A sample of 50 stocks traded on the NYSE that day showed that 24 went up. What is your point estimate of the population proportion of stocks that went up? c. Conduct your hypothesis test using \(\alpha=.01\) as the level of significance. What is your conclusion?

Short Answer

Expert verified
Reject the null hypothesis; there is evidence that the proportion is not 0.30.

Step by step solution

01

Formulate Hypotheses

The null hypothesis claims that the true proportion of stocks that increased in price is equal to the analyst's claim, i.e., 0.30. The alternative hypothesis suggests that the proportion is not 0.30.- Null Hypothesis \( H_0 \): \( p = 0.30 \)- Alternative Hypothesis \( H_a \): \( p eq 0.30 \)
02

Calculate Point Estimate

To find the point estimate for the population proportion, divide the number of stocks that went up by the total number of sampled stocks.\[p = \frac{24}{50} = 0.48\]Thus, the point estimate of the population proportion is 0.48.
03

Set Up the Hypothesis Test

Use a significance level of \( \alpha = 0.01 \). Since we are testing differences from the claimed proportion (two-tailed), compute the test statistic using the formula:\[z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}\]where \( \hat{p} = 0.48 \), \( p_0 = 0.30 \), and \( n = 50 \).
04

Calculate the Test Statistic

Plug the values into the formula:\[z = \frac{0.48 - 0.30}{\sqrt{\frac{0.30 \times (1-0.30)}{50}}}\]\[z = \frac{0.18}{\sqrt{\frac{0.21}{50}}} = \frac{0.18}{0.0648} \approx 2.78\]The test statistic, \( z \), is approximately 2.78.
05

Decision Rule

At a significance level of \( \alpha = 0.01 \), find the critical z-values for a two-tailed test. These are approximately \( \pm 2.58 \). If the test statistic exceeds these critical values in absolute value, we reject the null hypothesis.
06

Make a Decision

Since the calculated test statistic, 2.78, is greater than the critical value of 2.58, we reject the null hypothesis. There is sufficient evidence to suggest that the proportion of stocks that increased does not equal 0.30.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Null Hypothesis
In the realm of statistical hypothesis testing, the null hypothesis serves as the foundational assumption to be tested. It represents a statement of no effect or no difference. In our example, the null hypothesis posits that the proportion of stocks that increased in price on the specified day is exactly as claimed by the financial analyst: 0.30. This hypothesis can be crucial when making predictions or validating claims since it assumes that the observed effect is purely coincidental and not due to any true difference or change.

The null hypothesis is often symbolized as \[H_0: p = 0.30\] Here, \(p\) stands for the proportion of the total number of stocks that increased. An essential goal in hypothesis testing is to determine whether the data at hand provides sufficient evidence against the null hypothesis to reject it. By assuming that the observed data resulted from random chance, statistical tests can assess if the data significantly deviates from this assumption, guiding us to potentially reject the null hypothesis.
Alternative Hypothesis
While the null hypothesis suggests no difference, the alternative hypothesis argues for a change, difference, or effect that may exist. It represents the outcome that a researcher thinks is true or wants to prove. In the given example, the alternative hypothesis challenges the claim, implying that the proportion of stocks that went up on that day is not 0.30.

Symbolically, the alternative hypothesis is written as: \[H_a: p eq 0.30\] This is a two-tailed hypothesis because it anticipates that the proportion could be either less than or more than 0.30. Unlike the null hypothesis, the result of the alternative hypothesis indicates that there's an effect or a difference that the data suggests. It is the hypothesis that researchers or analysts try to gather support for, suggesting that there is significant evidence for a deviation from the null.
Significance Level
The significance level, often denoted by \(\alpha\), is a critical component in hypothesis testing. It defines the threshold for deciding whether to reject the null hypothesis. In essence, it sets the probability of making a Type I error, which occurs when the null hypothesis is wrongly rejected. In our case, the significance level is chosen as 0.01. This means there's a 1% risk of concluding that a difference exists when there is none.

When conducting a hypothesis test, we compare the test statistic to a critical value determined by this significance level. The decision rule follows: if the test statistic falls outside the critical value range, the null hypothesis is rejected. A smaller \(\alpha\) makes it more stringent, suggesting that stronger evidence is needed to reject the null hypothesis. By setting \(\alpha = 0.01\), it indicates that the analyst requires very strong evidence against the null hypothesis to consider it unlikely.
Test Statistic
A test statistic is a standardized value that results from the hypothesis test calculations. It helps determine the difference between observed data and what would be expected under the null hypothesis. Calculating the test statistic helps assess the evidence against the null hypothesis in a quantitative manner.

In our example, the z-test statistic formula is used, which measures how far the sample proportion is from the null hypothesis proportion in standard error units: \[z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}\] Here, \(\hat{p}\) is the observed sample proportion, \(p_0\) is the hypothesized proportion (0.30), and \(n\) is the sample size. A calculated z-value of approximately 2.78 suggests a noteworthy deviation from the null hypothesis value. When this z-value surpasses the critical value of \(\pm 2.58\) for a 0.01 significance level in a two-tailed test, it offers significant evidence that the null hypothesis may be false, leading to its rejection.

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Most popular questions from this chapter

A radio station in Myrtle Beach announced that at least \(90 \%\) of the hotels and motels would be full for the Memorial Day weekend. The station advised listeners to make reservations in advance if they planned to be in the resort over the weekend. On Saturday night a sample of 58 hotels and motels showed 49 with a no-vacancy sign and 9 with vacancies. What is your reaction to the radio station's claim after secing the sample evidence? Use \(a=.05\) in making the statistical test. What is the \(p\) -value?

The manager of the Danvers-Hilton Resort Hotel stated that the mean guest bill for a weekend is \(\$ 600\) or less. A member of the hotel's accounting staff noticed that the total charges for guest bills have been increasing in recent months. The accountant will use a sample of future weekend guest bills to test the manager's claim. a. Which form of the hypotheses should be used to test the manager's claim? Explain. \\[\begin{array}{lll}H_{0}: \mu \geq 600 & H_{0}: \mu \leq 600 & H_{0}: \mu=600 \\ H_{\mathrm{a}}: \mu<600 & H_{\mathrm{a}}: \mu>600 & H_{\mathrm{a}}: \mu \neq 600\end{array}\\] b. What conclusion is appropriate when \(H_{0}\) cannot be rejected? c. What conclusion is appropriate when \(H_{0}\) can be rejected?

Consider the following hypothesis test: \\[ \begin{array}{l} H_{0}: \mu=22 \\ H_{\mathrm{a}}: \mu \neq 22 \end{array} \\] A sample of 75 is used and the population standard deviation is \(10 .\) Compute the \(p\) -value and state your conclusion for each of the following sample results. Use \(\alpha=.01\) a. \(\bar{x}=23\) b. \(\bar{x}=25.1\) c. \(\quad \bar{x}=20\)

In a study entitled How Undergraduate Students Use Credit Cards, it was reported that undergraduate students have a mean credit card balance of \(\$ 3173\) (Sallie Mae, April 2009 ). This figure was an all-time high and had increased \(44 \%\) over the previous five years. Assume that a current study is being conducted to determine if it can be concluded that the mean credit card balance for undergraduate students has continued to increase compared to the April 2009 report. Based on previous studies, use a population standard deviation \(\sigma=\) \(\$ 1000\) a. State the null and alternative hypotheses. b. What is the \(p\) -value for a sample of 180 undergraduate students with a sample mean credit card balance of \(\$ 3325 ?\) c. Using a .05 level of significance, what is your conclusion?

A recent article concerning bullish and bearish sentiment about the stock market reported that \(41 \%\) of investors responding to an American Institute of Individual Investors (AAII) poll were bullish on the market and \(26 \%\) were bearish (USA Today, January 11,2010 ). The article also reported that the long-term average measure of bullishness is .39 or \(39 \%\) Suppose the AAII poll used a sample size of \(450 .\) Using .39 (the long-term average) as the population proportion of investors who are bullish, conduct a hypothesis test to determine if the current proportion of investors who are bullish is significantly greater than the longterm average proportion. a. State the appropriate hypotheses for your significance test. b. Use the sample results to compute the test statistic and the \(p\) -value. c. Using \(\alpha=.10,\) what is your conclusion?

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