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Consider using a z test to test \(H_{0}: p=.6 .\) Determine the \(P\)-value in each of the following situations. a. \(H_{a}: p>.6, z=1.47\) b. \(H_{\mathrm{a}}: p<.6, z=-2.70\) c. \(H_{a}: p \neq .6, z=-2.70\) d. \(H_{a}: p<.6, z=.25\)

Short Answer

Expert verified
(a) 0.0708, (b) 0.0035, (c) 0.0070, (d) 0.5987

Step by step solution

01

Understand the Z-Test

The Z-test is used to determine if there is a significant difference between a sample proportion and a known population proportion. In this case, the null hypothesis is that the population proportion is 0.6, and we are given a Z-score for each scenario.
02

Determine the Appropriate Alternative Hypothesis

For each situation, identify the alternative hypothesis: - (a) The alternative hypothesis is that the population proportion is greater than 0.6. - (b) The alternative hypothesis is that the population proportion is less than 0.6. - (c) The alternative hypothesis is that the population proportion is not equal to 0.6. - (d) The alternative hypothesis is that the population proportion is less than 0.6.
03

Find the P-Value for Each Scenario

To find the P-value, use the Z-score provided:- **(a)** For a one-tailed test where the alternative hypothesis is greater than, find the P-value as the area to the right of the Z-score. When \( z = 1.47 \), the P-value for \( H_a: p > 0.6 \) is found using \( 1 - \text{norm.cdf}(1.47) \).- **(b)** For a one-tailed test where the alternative hypothesis is less than, find the P-value as the area to the left of the Z-score. When \( z = -2.70 \), the P-value for \( H_a: p < 0.6 \) is \( \text{norm.cdf}(-2.70) \).- **(c)** For a two-tailed test, find the P-value as twice the area of the smaller tail. Since \( z = -2.70 \), the P-value is \( 2 \times \text{norm.cdf}(-2.70) \).- **(d)** With \( z = 0.25 \) and the alternative \( H_a: p < 0.6 \), the P-value is \( \text{norm.cdf}(0.25) \).
04

Calculate Each P-Value

Using the standard normal distribution table or a calculator, calculate:- **(a)** For \( z = 1.47 \), the area to the left is approximately 0.9292, leaving a P-value for \( p > 0.6 \) of \( 1 - 0.9292 = 0.0708 \).- **(b)** For \( z = -2.70 \), the area is approximately 0.0035.- **(c)** For the two-tailed test, this is \( 2 \times 0.0035 = 0.0070 \).- **(d)** For \( z = 0.25 \), the area to the left is approximately 0.5987.
05

Report Results

Concluding the calculations: - **(a)** The P-value is 0.0708 - **(b)** The P-value is 0.0035 - **(c)** The P-value is 0.0070 - **(d)** The P-value is 0.5987

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Key Concepts

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

hypothesis testing
Hypothesis testing is a statistical method that helps us decide whether to reject a null hypothesis based on sample data. It starts with formulating two opposite statements: the null hypothesis (\( H_0 \)) and the alternative hypothesis (\( H_a \)).
The null hypothesis represents the status quo or a specific condition to be tested. For instance, it might claim that a population parameter, such as a proportion, equals a certain value. The aim is to determine if there is enough evidence in the sample data to reject this assumption.
The alternative hypothesis, on the other hand, is what you might want to prove. If the sample data provides enough evidence against \( H_0 \), you would reject the null hypothesis in favor of \( H_a \).
When conducting hypothesis testing, a threshold called the significance level is used, often denoted by \( \alpha \). Common choices for \( \alpha \) are 0.05 or 0.01, which determine how much risk you are willing to take for rejecting a true null hypothesis. This process usually involves calculating a test statistic based on your sample data, such as a Z-score in the case of Z-tests. This statistic helps to identify the P-value which will guide the decision to accept or reject \( H_0 \).
p-value calculation
The P-value is a crucial element in hypothesis testing that helps determine the strength of the evidence in your sample data. It measures the probability of obtaining a test result as extreme, or more extreme, than the one observed, assuming the null hypothesis is true.
To calculate a P-value from a Z-test, you first need the Z-score, which is a standardized statistic indicating how many standard deviations your sample is away from the null hypothesis parameter value.
For a one-tailed test where \( H_a \) is greater than \( p \), the P-value is the area to the right of the Z-score. In contrast, if \( H_a \) is less than \( p \), the P-value is the area to the left. For two-tailed tests where \( p \) is not equal to a specific value, the P-value is twice the area of the smallest tail.
  • For example, if \( z = 1.47 \) and \( H_a: p > 0.6 \), find the P-value using \( 1 - \text{norm.cdf}(1.47) \).
  • If \( z = -2.70 \), and \( H_a: p < 0.6 \), the P-value is \( \text{norm.cdf}(-2.70) \).
  • For a two-tailed test with \( z = -2.70 \), compute \( 2 \times \text{norm.cdf}(-2.70) \).
Finding the P-value helps you determine whether your sample data is unusual under the null hypothesis, guiding you in making the decision to reject or not reject \( H_0 \).
normal distribution
The normal distribution is a fundamental concept in statistics and vital for understanding hypothesis testing. Often depicted as a bell-shaped curve, it describes how data tends to be distributed around a central value.
In a normal distribution:
  • The mean, median, and mode are all equal.
  • The curve is symmetrical around the mean.
  • Standard deviation dictates the width of the bell curve: larger values lead to a wider curve.
A standard normal distribution is a special case with a mean of 0 and a standard deviation of 1, often used in Z-tests. The Z-score, used in hypothesis testing, transforms the data into this standardized form, allowing easy calculation of probabilities (P-values) and comparison across various datasets.
Understanding normal distribution aids in assessing how typical or atypical your sample data is. It allows you to find areas under the curve (probabilities) related to Z-scores, which is crucial for determining P-values in hypothesis testing.
alternative hypothesis
The alternative hypothesis (\( H_a \)) is a fundamental part of hypothesis testing. It is the statement you aim to support with your sample data. If sufficient evidence exists, you will reject the null hypothesis in favor of the alternative hypothesis.
The alternative hypothesis can be one of the following types depending on the research question:
  • One-tailed: This considers only one direction of effect. For instance, stating \( H_a: p > 0.6 \) suggests that the population proportion is greater than a certain value.
  • Two-tailed: This looks at deviations in both directions. \( H_a: p eq 0.6 \) implies that the population proportion might be either less than or greater than this value.
Choosing the correct alternative hypothesis is vital, as it defines how you will compute P-values and interpret your statistical outcomes. In hypothesis testing, the goal is to assess whether the observed data provides enough evidence to support \( H_a \), making it essential to clearly define the alternative hypothesis from the beginning of your analysis.

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Most popular questions from this chapter

Lightbulbs of a certain type are advertised as having an average lifetime of 750 hours. The price of these bulbs is very favorable, so a potential customer has decided to go ahead with a purchase arrangement unless it can be conclusively demonstrated that the true average lifetime is smaller than what is advertised. A random sample of 50 bulbs was selected, the lifetime of each bulb determined, and the appropriate hypotheses were tested using Minitab, resulting in the accompanying output. \(\begin{array}{lrrrrrr}\text { Variable } & N & \text { Mean } & \text { StDev } & \text { SE Mean } & \text { Z } & \text { P-Value } \\ \text { lifetime } 50 & 738.44 & 38.20 & 5.40 & -2.14 & 0.016\end{array}\) What conclusion would be appropriate for a significance level of \(.05\) ? A significance level of .01? What significance level and conclusion would you recommend?

The accompanying data on cube compressive strength (MPa) of concrete specimens appeared in the article "Experimental Study of Recycled Rubber-Filled High- Strength Concrete" (Magazine of Concrete \(\operatorname{Res}_{*}\) 2009: 549-556): \(\begin{array}{rrrrr}112.3 & 97.0 & 92.7 & 86.0 & 102.0 \\ 99.2 & 95.8 & 103.5 & 89.0 & 86.7\end{array}\) a. Is it plausible that the compressive strength for this type of concrete is normally distributed? b. Suppose the concrete will be used for a particular application unless there is strong evidence that true average strength is less than 100 MPa. Should the concrete be used? Carry out a test of appropriate hypotheses.

A manufacturer of nickel-hydrogen batteries randomly selects 100 nickel plates for test cells, cycles them a specified number of times, and determines that 14 of the plates have blistered. a. Does this provide compelling evidence for concluding that more than \(10 \%\) of all plates blister under such circumstances? State and test the appropriate hypotheses using a significance level of \(.05\). In reaching your conclusion, what type of error might you have committed? b. If it is really the case that \(15 \%\) of all plates blister under these circumstances and a sample size of 100 is used, how likely is it that the null hypothesis of part (a) will not be rejected by the level .05 test? Answer this question for a sample size of \(200 .\) c. How many plates would have to be tested to have \(\beta(.15)=.10\) for the test of part (a)?

A mixture of pulverized fuel ash and Portland cement to be used for grouting should have a compressive strength of more than \(1300 \mathrm{KN} / \mathrm{m}^{2}\). The mixture will not be used unless experimental evidence indicates conclusively that the strength specification has been met. Suppose compressive strength for specimens of this mixture is normally distributed with \(\sigma=60\). Let \(\mu\) denote the true average compressive strength. a. What are the appropriate null and alternative hypotheses? b. Let \(\bar{X}\) denote the sample average compressive strength for \(n=10\) randomly selected specimens. Consider the test procedure with test statistic \(\bar{X}\) itself (not standardized). If \(\bar{x}=1340\), should \(H_{0}\) be rejected using a significance level of \(.01\) ? [Hint: What is the probability distribution of the test statistic when \(H_{0}\) is true?] c. What is the probability distribution of the test statistic when \(\mu=1350\) ? For a test with \(\alpha=.01\), what is the probability that the mixture will be judged unsatisfac- tory when in fact \(\mu=1350\) (a type II error)?

The recommended daily dietary allowance for zinc among males older than age 50 years is \(15 \mathrm{mg} /\) day. The article "Nutrient Intakes and Dietary Patterns of Older Americans: A National Study" \((J .\) of Gerontology, 1992: M145-150) reports the following summary data on intake for a sample of males age 65-74 years: \(n=115, \bar{x}=11.3\), and \(s=6.43\). Does this data indicate that average daily zinc intake in the population of all males ages \(65-74\) falls below the recommended allowance?

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