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A true/false test has 50 questions. Suppose a passing grade is 35 or more correct answers. Test the claim that a student knows more than half of the answers and is not just guessing. Assume the student gets 35 answers correct out of \(50 .\) Use a significance level of \(0.05 .\) Steps 1 and 2 of a hypothesis test procedure are given. Show steps 3 and 4, and be sure to write a clear conclusion. $$ \text { Step 1: } \begin{aligned} &\mathrm{H}_{0}: p=0.50 \\ &\mathrm{H}_{\mathrm{a}}: p>0.50 \end{aligned} $$ Step 2: Choose the one-proportion z-test. Sample size is large enough, because \(n p_{0}\) is \(50(0.5)=25\) and \(n\left(1-p_{0}\right)=50(0.50)=25\), and both are more than \(10 .\) Assume the sample is random and \(\alpha=0.05\).

Short Answer

Expert verified
The student seems to know more than half of the answers since we are rejecting the null hypothesis at a 0.05 significance level.

Step by step solution

01

Calculate the Test Statistic

Here we need to calculate the z-score to test the student's score against the proportion under null hypothesis which is 0.50. The z-score is calculated as follows: \( z = \frac{(p - p_{0})}{\sqrt{(p_{0}(1 - p_{0})/n)}} \) where \( p = x/n \) is the sample proportion, \( p_{0} = 0.50 \)is the assumed population proportion under the null hypothesis, and n is the number of questions (here n=50). By substituting the values we get \( z = \frac{(0.70 - 0.50)}{\sqrt{(0.50(1 - 0.50)/50)}} \) After calculation we get z approximately equal to 4.472.
02

Determine the Critical Z-value and Compare with Test Statistic

The critical z-value to the right of the mean at 0.05 significance level is 1.645 (you can refer to z-table for this). If the calculated z-value falls in the rejection region we reject the null hypothesis in favor of alternative. Since 4.472 > 1.645, the observed z-value falls in the rejection region.
03

Write the Conclusion

Since the test statistic is in the rejection region, we reject the null hypothesis. This implies that we have enough evidence at the 0.05 significance level to support the claim that the student knows more than half of the answers and not just guessing.

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Key Concepts

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

Z-test
The z-test is a statistical method used to determine whether there is a significant difference between sample observations and a known parameter of the population. When students face a problem involving hypothesis testing for proportions, a z-test is particularly useful if the sample size is large enough, usually defined as having at least 10 successes and 10 failures in the sample (as per the Central Limit Theorem).

In the educational example provided, a z-test is the appropriate choice to assess whether a student's test result is significantly greater than mere chance. Calculation of the z-score involves comparing the sample proportion to the hypothesized population proportion, accounting for the sample size. In essence, the z-score tells us how many standard deviations away the sample's proportion is from the expected population proportion.
Statistical Significance
Statistical significance is a term used to convey that the observed effect in the data is unlikely due to chance alone. This is evaluated against a pre-determined threshold called the significance level, notated as alpha (\r\text{\alpha}\r\text{\(\)}).

In our exercise, the significance level is set at 0.05. This means that there is a 5% risk of concluding that an effect or difference exists when there is none (Type I error). By comparing the calculated z-score against a critical value obtained from the z-table, researchers decide whether to reject the null hypothesis. If the z-score exceeds the critical value, the result is statistically significant, indicating strong evidence against the null hypothesis. Thus, the insignificance level is a guardrail for making informed decisions based on data.
Null Hypothesis
The null hypothesis (\r\text{\(H_0\)}\r\text{\(\)}) represents the default statement or position that there is no effect or no difference. It serves as a benchmark for testing whether the evidence from the data is strong enough to warrant a conclusion that there's something noteworthy happening.

In the context of our textbook problem, the null hypothesis posits that the student's knowledge is equivalent to guessing, with the probability (p) of a correct answer being 0.50 for each question. The null hypothesis always specifies an exact value, which in hypothesis testing, is what we attempt to challenge with the evidence from the sample.
Alternative Hypothesis
Conversely to the null, the alternative hypothesis (\r\text{\(H_a\)}\r\text{\(\)}) is the claim that there is an effect or a difference, or specifically, that the observed data is caused by something other than chance. It's the hypothesis researchers typically aim to support.

In our example, the alternative hypothesis claims that the student knows more than half of the answers (\r\text{\(p > 0.50\)}\r\text{\(\)}). Notice it is stated in terms of an inequality, which makes this a one-tailed test, looking for evidence of an effect in one direction (in this case, greater than the null hypothesis value).
Test Statistic
A test statistic is a value calculated from the sample data that is used to assess the veracity of the null hypothesis. It synthesizes the sample information into a single number that can be used to determine how consistent the sample is with the null hypothesis.

In hypothesis testing, whether employing a z-test, t-test, or any other test, the test statistic is what we compare against critical values to decide on the null hypothesis. In our exercise, the z-score of approximately 4.472 is a test statistic that, when compared to the critical value, suggested a significant result, hence we rejected the null hypothesis, concluding the student did in fact possess knowledge beyond guessing.

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Most popular questions from this chapter

Biased Coin? A study is done to see whether a coin is biased. The alternative hypothesis used is two-sided, and the obtained z-value is 2. Assuming that the sample size is sufficiently large and that the other conditions are also satisfied, use the Empirical Rule to approximate the p-value.

A proponent of a new proposition on a ballot wants to know the population percentage of people who support the bill. Suppose a poll is taken, and 580 out of 1000 randomly selected people support the proposition. Should the proponent use a hypothesis test or a confidence interval to answer this question? Explain. If it is a hypothesis test, state the hypotheses and find the test statistic, p-value, and conclusion. Use a \(5 \%\) significance level. If a confidence interval is appropriate, find the approximate \(95 \%\) confidence interval. In both cases, assume that the necessary conditions have been met.

Refer to Exercise \(8.97 .\) Suppose 14 out of 20 voters in Pennsylvania report having voted for an independent candidate. The null hypothesis is that the population proportion is \(0.50 .\) What value of the test statistic should you report?

Historically (from about 2001 to 2014 ), \(57 \%\) of Americans believed that global warming is caused by human activities. A March 2017 Gallup poll of a random sample of 1018 Americans found that 692 believed that global warming is caused by human activities. a. What percentage of the sample believed global warming was caused by human activities? b. Test the hypothesis that the proportion of Americans who believe global warming is caused by human activities has changed from the historical value of \(57 \%\). Use a significance level of \(0.01\). c. Choose the correct interpretation: i. In 2017 , the percentage of Americans who believe global warming is caused by human activities is not significantly different from \(57 \%\). ii. In 2017 , the percentage of Americans who believe global warming is caused by human activities has changed from the historical level of \(57 \%\).

If we reject the null hypothesis, can we claim to have proved that the null hypothesis is false? Why or why not?

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