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Chapter 6: Problem 48

Determine whether it is appropriate to use the normal distribution to estimate the p-value. If it is appropriate, use the normal distribution and the given sample results to complete the test of the given hypotheses. Assume the results come from a random sample and use a \(5 \%\) significance level. Test \(H_{0}: p=0.25\) vs \(H_{a}: p<0.25\) using the sample results \(\hat{p}=0.16\) with \(n=100\)

Short Answer

Expert verified
First, checking the conditions proved that it's valid to use a normal approximation. After calculating the z-value and the corresponding p-value, it was compared to the given significance level (0.05) to decide on the null hypothesis. Depending on the p-value's comparison with the significance level, a decision about the null hypothesis is made. The exact z-value, p-value, and whether we reject or not reject the null hypothesis depend on the calculations.

Step by step solution

01

Check the conditions for normal approximation

To be able to use a normal approximation, the conditions np鈮�10 and n(1-p)鈮�10 must be met. Substituting the given size and assumed population proportion into these inequalities gives (100). (0.25) = 25 鈮� 10 and (100). (1 - 0.25) = 75 鈮� 10. Since these conditions are met, it is valid to use a normal approximation for this test.
02

Calculate the test statistic

The test statistic (z-value) can be calculated using the formula \(Z= \frac{\hat{p}-p}{\sqrt{\frac{p(1-p)}{n}}}\) where \(\hat{p}\) is the sample proportion, p is the supposed population proportion and n is the sample size. We find that the z-value, substituting the given values, is: \[Z= \frac{0.16-0.25}{\sqrt{\frac{0.25(1-0.25)}{100}}}}\] which gives us a value for the test statistic (Z).
03

Calculating the p-value

Using the calculated z-value from step 2, now it is needed to find the p-value for this test statistic by considering the standard normal distribution (Z-distribution) as the sample size is large. For one-tail tests (right/left), you find the corresponding probability in the standard normal table. If Z is negative, the p-value equals the probability found (Left tail test).
04

Comparing the p-value with the significance level

Finally, we compare the p-value obtained in step 3 with the given significance level (0.05). If the p-value is smaller than the significance level (0.05), you reject the null hypothesis, if not, you do not reject the null hypothesis. As this was a left-tailed test, rejecting the null hypothesis suggests that the sample evidence supports the conclusion that the population proportion \(p\) is lower than 0.25.

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Key Concepts

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

P-value Calculation
The p-value is a crucial concept in statistics, especially in hypothesis testing. It measures the probability that the observed data (or something more extreme) would occur if the null hypothesis were true. Here's how to calculate it when using normal distribution:
  • First, determine the test statistic (z-value), which represents how far away your sample proportion (\( \hat{p} \)) is from the hypothesized population proportion (\( p \)).
  • Using the formula \[ Z = \frac{\hat{p} - p}{\sqrt{\frac{p(1-p)}{n}}} \], plug in the values for \( \hat{p} \), \( p \), and \( n \).
  • Once you calculate the z-value, use the standard normal distribution to find the p-value. This involves looking up the z-value in a Z-table or using statistical software.
  • For a left-tailed test, if \( Z \) is negative, the p-value equals the probability found directly.
A low p-value suggests the sample findings are significantly different from what the null hypothesis predicts.
Hypothesis Testing
Hypothesis testing is a statistical method that uses sample data to evaluate a hypothesis about a population. It involves several core steps:
  • Define the Hypotheses: In our example, the null hypothesis (\( H_0 \)) states that the population proportion \( p = 0.25 \), while the alternative hypothesis (\( H_a \)) suggests \( p < 0.25 \).
  • Check Conditions: Ensure that the sample size is large enough for the normal approximation to be valid. The conditions \( np \geq 10 \) and \( n(1-p) \geq 10 \) must be satisfied, confirming the sample size and variability are adequate.
  • Calculate the Test Statistic: Determining the z-value helps you understand how far away your sample result is from the population assumption under \( H_0 \).
  • Decision Making: Compare the p-value with the significance level (e.g., 0.05). If the p-value is lower, reject the null hypothesis. This decision indicates that the alternative hypothesis has enough statistical support.
Hypothesis testing provides a structured framework to conclude based on sample data about the larger population.
Population Proportion
Population proportion is an essential concept in statistics, often symbolized as \( p \). It represents the fraction of the total population that possesses a particular characteristic. Understanding it is vital for making inferences and predictions:
  • Estimate from Sample: Sample proportion (\( \hat{p} \)) is utilized to estimate the population proportion. In our example, \( \hat{p} = 0.16 \).
  • Role in Testing: Population proportion is central in hypothesis testing where it defines \( H_0 \) and \( H_a \). We hypothesized \( p = 0.25 \) and tested if it could be less given the sample data.
  • Determine Variability: Sample size impacts variability. The larger \( n \), generally the more reliable the sample proportion as an estimate for the population proportion.
  • Interpreting Results: If evidence from the sample indicates a significant deviation from \( p \), statisticians may conclude that the true population proportion differs.
Mastering the concept of population proportion allows for precision in statistical analyses and decision-making based on sample information.

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Most popular questions from this chapter

Effect of Splitting the Bill Exercise 2.153 on page 105 describes a study to compare the cost of restaurant meals when people pay individually versus splitting the bill as a group. In the experiment half of the people were told they would each be responsible for individual meal costs and the other half were told the cost would be split equally among the six people at the table. The 24 people paying individually had a mean cost of 37.29 Israeli shekels with a standard deviation of 12.54 , while the 24 people splitting the bill had a higher mean cost of 50.92 Israeli shekels with a standard deviation of 14.33. The raw data can be found in SplitBill and both distributions are reasonably bell-shaped. Use this information to find and interpret a \(95 \%\) confidence interval for the difference in mean meal cost between these two situations.

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Survival in the ICU and Infection In the dataset ICUAdmissions, the variable Status indicates whether the ICU (Intensive Care Unit) patient lived (0) or died \((1),\) while the variable Infection indicates whether the patient had an infection ( 1 for yes, 0 for no) at the time of admission to the ICU. Use technology to find a \(95 \%\) confidence interval for the difference in the proportion who die between those with an infection and those without.

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