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For the hypothesis test \(H_{0}: \mu=5\) against \(H_{1}: \mu<5\) with variance unknown and \(n=12\), approximate the \(P\) -value for each of the following test statistics. (a) \(t_{0}=2.05\) (b) \(t_{0}=-1.84\) (c) \(t_{0}=0.4\)

Short Answer

Expert verified
(a) P-value is large; (b) P-value ≈ 0.05 to 0.07; (c) P-value ≈ 0.65.

Step by step solution

01

Understand the hypothesis test

We are performing a one-sample t-test where the null hypothesis is \( H_0: \mu = 5 \) and the alternative hypothesis is \( H_1: \mu < 5 \). Since variance is unknown and sample size \( n = 12 \), we will use the t-distribution with \( n-1 = 11 \) degrees of freedom.
02

Identify the t-statistic and sample size

For each case, the test statistic \( t_0 \) is provided: (a) \( t_0 = 2.05 \), (b) \( t_0 = -1.84 \), (c) \( t_0 = 0.4 \). The sample size \( n = 12 \), which results in \( 11 \) degrees of freedom for the t-test.
03

Approximate the P-value for (a)

Given \( t_0 = 2.05 \), we need to find \( P(T_{11} < 2.05) \), but since the hypothesis is \( \mu < 5 \), we look for \( P(T_{11} < 2.05) \) which is large, so the \( P \)-value will be high, likely near 1. However, it's important to know that this P-value is not relevant since it does not support \( H_1 \).
04

Approximate the P-value for (b)

Given \( t_0 = -1.84 \), we need to find \( P(T_{11} < -1.84) \). Using a t-distribution table or a calculator, \( P(T_{11} < -1.84) \) can be found, which would normally be a value such as 0.05 to 0.07, indicating a moderate level of significance.
05

Approximate the P-value for (c)

Given \( t_0 = 0.4 \), we need to find \( P(T_{11} < 0.4) \). Checking using a t-distribution table, \( P(T_{11} < 0.4) \) would be around 0.65, which is a sizable P-value, not supporting \( H_1 \).

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Key Concepts

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

t-distribution
When dealing with small sample sizes and an unknown population variance, like in the example provided, the t-distribution is the tool to use for analyzing data. The t-distribution is similar to the normal distribution, but with fatter tails. This means it is more forgiving of outliers and provides a more realistic picture of the data's variability.
Unlike the normal distribution, the shape of the t-distribution depends on the degrees of freedom, becoming wider with fewer degrees of freedom and more similar to the normal distribution as degrees of freedom increase. In our example, with a sample size of 12, we have 11 degrees of freedom (sample size minus one). Knowing how the distribution works helps us understand where most of our data will fall and assists in calculating probabilities, known as P-values.
one-sample t-test
A one-sample t-test is a hypothesis test used to determine whether the mean of a single sample differs significantly from a known or hypothesized population mean. In our situation, we are testing the hypothesis that the sample mean is equal to 5 against the alternative that it is less than 5.
The choice to use a t-test arises because the population variance is unknown, and the sample size is relatively small (less than 30). This makes the t-distribution a suitable model for handling the variability in the data. During the test, we calculate the test statistic, often denoted as \( t_0 \). This value gauges how far the sample mean deviates from the hypothesized population mean under the null hypothesis.
P-value calculation
The P-value is a critical aspect of hypothesis testing as it helps you determine the significance of your results. It represents the probability of obtaining a test statistic as extreme as the sample result, assuming the null hypothesis is true.
In the given problem, different P-values are calculated for different test statistics based on the t-distribution with 11 degrees of freedom. Here's a brief look at each:
  • For \( t_0 = 2.05 \), you determine \( P(T_{11} < 2.05) \), which would be quite large. However, because this P-value does not align with the \( H_1: \mu < 5 \), it indicates that the null hypothesis holds.
  • For \( t_0 = -1.84 \), \( P(T_{11} < -1.84) \) indicates a moderate level of significance, often enough to consider supporting \( H_1 \).
  • For \( t_0 = 0.4 \), \( P(T_{11} < 0.4) \) results in a substantial P-value, suggesting that you cannot reject the null hypothesis.
A smaller P-value suggests stronger evidence against the null hypothesis, leading you towards considering the alternative hypothesis as the more likely scenario.
degrees of freedom
Degrees of freedom are a concept essential to various statistical computations, including the t-test. They refer to the number of values that can vary independently in your calculation once you've considered the required fixed parameters, like sample mean or population mean.
In a one-sample t-test, the degrees of freedom are calculated as the sample size minus one, namely \( n-1 \). For our example, with a sample size \( n = 12 \), the degrees of freedom are \( 12 - 1 = 11 \). Understanding degrees of freedom is important because it influences how the t-distribution is used. As degrees of freedom increase, the t-distribution progressively approaches the standard normal distribution.
Essentially, degrees of freedom help to tailor your test statistic to the sample size and data variability, ensuring that the inference you draw from the test is reliable and accurate given the context of the data you have.

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Most popular questions from this chapter

A new type of tip can be used in a Rockwell hardness tester. Eight coupons from test ingots of a nickel-based alloy are selected, and each coupon is tested using the new tip. The Rockwell C-scale hardness readings are 63,65,58,60 \(55,57,53,\) and \(59 .\) Do the results support the claim that the mean hardness exceeds 60 at a 0.05 level?

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The impurity level (in ppm) is routinely measured in an intermediate chemical product. The following data were observed in a recent test: $$ \begin{array}{l} 2.4,2.5,1.7,1.6,1.9,2.6,1.3,1.9,2.0,2.5,2.6,2.3,2.0,1.8, \\ 1.3,1.7,2.0,1.9,2.3,1.9,2.4,1.6 \end{array} $$ Can you claim that the median impurity level is less than \(2.5 \mathrm{ppm} ?\) (a) State and test the appropriate hypothesis using the sign test with \(\alpha=0.05 .\) What is the \(P\) -value for this test? (b) Use the normal approximation for the sign test to test \(H_{0}: \widetilde{\mu}=2.5\) versus \(H_{1}: \tilde{\mu}<2.5 .\) What is the \(P\) -value for this test?

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