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For the hypothesis test \(H_{0}: \mu=7\) against \(H_{1}: \mu \neq 7\) with variance unknown and \(n=20\), 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 ≈ 0.05, (b) P-value ≈ 0.082, (c) P-value ≈ 0.68

Step by step solution

01

Understanding the problem

We need to find the P-value for a t-test where the null hypothesis is that the mean \( \mu = 7 \) and the alternative is that \( \mu eq 7 \), using a t-statistic with 19 degrees of freedom (since \( n = 20 \)). We have three different test statistics to evaluate.
02

Determine degrees of freedom

The degrees of freedom for a t-distribution when variance is unknown is given by \( n - 1 \). Since \( n = 20 \), the degrees of freedom is \( n - 1 = 19 \).
03

P-value for (a) \( t_0 = 2.05 \)

For \( t_0 = 2.05 \) with 19 degrees of freedom, we will use a t-distribution table or calculator. The P-value equals two times the area to the right of \( |2.05| \). Look up \( t = 2.05 \) in the t-table or use a calculator, which gives an area of roughly 0.025 in one tail. Therefore, the two-tailed P-value is approximately 0.05.
04

P-value for (b) \( t_0 = -1.84 \)

For \( t_0 = -1.84 \), calculate the two-tailed P-value. Look for \( t = 1.84 \) with 19 degrees of freedom. The area to the right of \( 1.84 \) is about 0.041. Thus, the two-tailed P-value is approximately \( 2 \times 0.041 = 0.082 \).
05

P-value for (c) \( t_0 = 0.4 \)

For \( t_0 = 0.4 \), check the t-table or use a calculator to find the area for \( t = 0.4 \) with 19 degrees of freedom. The P-value will be very high, typically 0.34 for one tail. Thus, the two-tailed P-value is approximately 0.68.

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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 method to make decisions in statistics about the population based on sample data. It starts with setting up two competing hypotheses, the null hypothesis (H_{0}) and the alternative hypothesis (H_{1}). The null hypothesis is often a statement of no effect or no difference; in our case, it posits that the population mean \( \mu \) is 7. The alternative hypothesis represents the opposite, suggesting that the population mean is not 7. This is written as \( \mu eq 7 \).
  • Null Hypothesis (H_{0}): \( \mu = 7 \).
  • Alternative Hypothesis (H_{1}): \( \mu eq 7 \).
Hypothesis testing involves comparing sample data against these hypotheses. By analyzing the data, often through a test statistic, we determine whether there's enough evidence to reject the null hypothesis. A t-test is used when the variance is unknown and the sample size is small, and it helps check how much the sample mean differs from a hypothesized population mean.
Degrees of Freedom
Degrees of freedom (df) in statistics describe the number of values in a calculation that are free to vary. When conducting a t-test for hypothesis testing, especially when variances are unknown, determining the degrees of freedom is a crucial step. This concept helps outline how "spread out" the distributions will be. When utilizing the t-distribution, degrees of freedom are calculated based on the sample size. The formula \( df = n - 1 \) means that our degrees of freedom are one less than the sample size. For instance, in the problem given, a sample size of 20 (\( n = 20 \)) means the degrees of freedom are: \( df = 20 - 1 = 19 \).
  • This calculation is important because it affects the shape of the t-distribution used in hypothesis testing.
  • More degrees of freedom generally result in a distribution that closely resembles a normal distribution.
P-Value Calculation
Calculating the P-value is a critical part of hypothesis testing. The P-value tells you how extreme the data is. It measures the probability of observing a test statistic as extreme as, or more extreme than, the observed value under the assumption that the null hypothesis is true.In practical terms, for a t-test:
  • We determine the P-value by looking up the test statistic in a t-distribution table with the relevant degrees of freedom.
  • For a two-tailed test, as in our provided exercise scenario, the P-value is the area beyond the absolute value of the test statistic multiplied by two.
For example, in the exercise:
  • For \( t_{0} = 2.05 \), a lookup (or calculator) indicates the area to the right is roughly 0.025, and thus the two-tailed P-value is approximately 0.05.
  • For \( t_{0} = -1.84 \), the area on one side is about 0.041, giving a two-tailed P-value of 0.082.
  • For \( t_{0} = 0.4 \), which represents a smaller deviation from the mean, the P-value is much higher (about 0.68), indicating weaker evidence against H_{0}.
A small P-value suggests that the observed data are inconsistent with the null hypothesis, leading to its rejection in favor of the alternative hypothesis.

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Most popular questions from this chapter

An article in Growth: \(A\) Journal Devoted to Problems of Normal and Abnormal Growth ["Comparison of Measured and Estimated Fat-Free Weight, Fat, Potassium and Nitrogen of Growing Guinea Pigs" (Vol. \(46,\) No. \(4,1982,\) pp. \(306-321\) ) reported the results of a study that measured the body weight (in grams) for guinea pigs at birth. $$ \begin{array}{rrrrr} 421.0 & 452.6 & 456.1 & 494.6 & 373.8 \\ 90.5 & 110.7 & 96.4 & 81.7 & 102.4 \\ 241.0 & 296.0 & 317.0 & 290.9 & 256.5 \\ 447.8 & 687.6 & 705.7 & 879.0 & 88.8 \\ 296.0 & 273.0 & 268.0 & 227.5 & 279.3 \\ 258.5 & 296.0 & & & \end{array} $$ (a) Test the hypothesis that mean body weight is 300 grams. Use \(\alpha=0.05\) (b) What is the smallest level of significance at which you would be willing to reject the null hypothesis? (c) Explain how you could answer the question in part (a) with a two-sided confidence interval on mean body weight.

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