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The following hypotheses are given. $$ \begin{array}{l} H_{0}: \pi \leq .70 \\ H_{1}: \pi>.70 \end{array} $$ A sample of 100 observations revealed that \(p=.75 .\) At the .05 significance level, can the null hypothesis be rejected? a. State the decision rule. b. Compute the value of the test statistic. c. What is your decision regarding the null hypothesis?

Short Answer

Expert verified
Do not reject the null hypothesis; the test statistic is less than the critical value.

Step by step solution

01

Define the Decision Rule

We have to determine the decision rule using the critical value approach. Since this is a right-tailed test with a significance level of 0.05, we first look up the critical value in the z-table that corresponds to an area of 1 - 0.05 = 0.95. The critical z-value is 1.645. The decision rule is: reject the null hypothesis if the test statistic is greater than 1.645.
02

Compute the Test Statistic

The test statistic for a proportion is calculated using the formula:\[ z = \frac{p - \pi_0}{\sqrt{\frac{\pi_0(1-\pi_0)}{n}}} \]where \(p = 0.75\) is the sample proportion, \(\pi_0 = 0.70\) is the hypothesized population proportion, and \(n = 100\) is the sample size.Substitute in the values:\[ z = \frac{0.75 - 0.70}{\sqrt{\frac{0.70(1-0.70)}{100}}} = \frac{0.05}{0.0458} \approx 1.09\]Thus, the calculated value of the test statistic is approximately 1.09.
03

Make Decision Regarding the Null Hypothesis

Compare the calculated test statistic to the critical value determined in Step 1. The calculated z-value is 1.09, which is less than 1.645. Hence, we do not reject the null hypothesis, as the test statistic does not exceed the critical value.

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Key Concepts

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

Understanding the Null Hypothesis
In hypothesis testing, the null hypothesis is a crucial component. It's the initial statement about a population parameter, often suggesting there is no effect or no difference. In our example, the null hypothesis, denoted as \(H_0\), asserts that the population proportion \(\pi\) is less than or equal to 0.70.

The null hypothesis acts as a default position that researchers aim to test against. It is considered true until evidence suggests otherwise. This brings significance to conducting the test as you’re essentially challenging the status quo proposition about your data.

Key points about the null hypothesis include:
  • It is usually stated with the \(\leq\), \(=\), or \(\geq\) symbols.
  • The goal is to determine if there is significant evidence to reject this assumption.
Interpreting the null hypothesis correctly sets the stage for a successful hypothesis test.
Decoding the Critical Value
The critical value is an essential part of hypothesis testing. It helps determine the threshold at which you would reject or fail to reject the null hypothesis. Think of it as the line in the sand for decision making. This value is tied directly to the significance level, \(\alpha\), set by the researcher, which defines the risk of incorrectly rejecting the null hypothesis.

In the case study, a significance level of 0.05 was chosen, indicating a 5% risk level. This corresponds to a critical z-value obtained from a standard normal distribution table. Specifically, for a right-tailed test, a significance level of 0.05 gives a critical z-value of 1.645.

Checklist for critical value assignment:
  • Choose an appropriate significance level (common choices are 0.05, 0.01).
  • Decide if your test is one-tailed or two-tailed, which affects the look-up process in z-tables.
  • Use z-tables or t-tables to find the corresponding critical value.
Aligning the critical value with the test statistic helps determine the final decision on the hypothesis test.
Deriving the Test Statistic
The test statistic is a standardized value that helps compare observed data with the expected data under the null hypothesis. It effectively quantifies how far the sample proportion deviates from the population proportion stated in the null hypothesis.

In proportion hypothesis testing, especially with large sample sizes, we use the z-test statistic. The formula provided is:
- \[ z = \frac{p - \pi_0}{\sqrt{\frac{\pi_0(1-\pi_0)}{n}}} \]

It tells you how many standard deviations the sample proportion \(p\) is from the hypothesized proportion \(\pi_0\). For the exercise given, - the sample proportion \(p\) was 0.75,- the hypothesized population proportion was 0.70,- and the sample size \(n\) was 100.

Plug these into the formula to find a z-value of approximately 1.09. This computed test statistic is then compared to the critical value to make a decision.
Proportion Hypothesis Test
The proportion hypothesis test is specifically used to evaluate claims about a population proportion. It is particularly useful when dealing with binary outcomes like yes/no, pass/fail, etc.

Steps involved in carrying out a proportion hypothesis test include:
  • Formulate the null and alternative hypotheses, clearly stating the population proportion assumptions.
  • Calculate the test statistic using the sample data and the assumed null hypothesis proportion.
  • Determine the critical value based on your specified significance level.
  • Compare the test statistic with the critical value to draw a conclusion.
In the exercise presented, the null hypothesis stated that the population proportion \(\pi\) should be 0.70 or lower. After conducting the test, our test statistic 1.09 was less than the critical value 1.645, thus leading us not to reject the null hypothesis.

This sequence shows a careful balance between sample evidence and statistical benchmarks, forming the basis of decision-making in statistical practice.

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Most popular questions from this chapter

Tina Dennis is the comptroller for Meek Industries. She believes that the current cashflow problem at Meek is due to the slow collection of accounts receivable. She believes that more than \(60 \%\) of the accounts are more than 3 months in arrears. A random sample of 200 accounts showed that 140 were more than 3 months old. At the .01 significance level, can she conclude that more than \(60 \%\) of the accounts are in arrears for more than three months?

An urban planner claims that, nationally, \(20 \%\) of all families renting condominiums move during a given year. A random sample of 200 families renting condominiums in the Dallas Metroplex revealed that 56 moved during the past year. At the .01 significance level, does this evidence suggest that a larger proportion of condominium owners moved in the Dallas area? Determine the \(p\) -value.

After a losing season, there is a great uproar to fire the head football coach. In a random sample of 200 college alumni, 80 favor keeping the coach. Test at the .05 level of significance whether the proportion of alumni who support the coach is less than \(50 \%\).

A study was conducted to determine if there was a difference in the humor content in British and American trade magazine advertisements. In an independent random sample of 270 American trade magazine advertisements, 56 were humorous. An independent random sample of 203 British trade magazines contained 52 humorous ads. Do these data provide evidence at the .05 significance level that there is a difference in the proportion of humorous ads in British versus American trade magazines?

GfK Research North America conducted identical surveys 5 years apart. One question asked of women was "Are most men basically kind, gentle, and thoughtful?" The earlier survey revealed that, of the 3,000 women surveyed, 2,010 said that they were. The later survey revealed 1,530 of the 3,000 women thought that men were kind, gentle, and thoughtful. At the .05 level, can we conclude that women think men are less kind, gentle, and thoughtful in the later survey compared with the earlier one?

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