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Find the p-value. An independent random sample is selected from an approximately normal population with an unknown standard deviation. Find the p-value for the given set of hypotheses and \(T\) test statistic. Also determine if the null hypothesis would be rejected at \(\alpha=0.05\). (a) \(H_{A}: \mu>\mu_{0}, n=11, T=1.91\) (c) \(H_{A}: \mu \neq \mu_{0}, n=7, T=0.83\) (b) \(H_{A}: \mu<\mu_{0}, n=17, T=-3.45\) (d) \(H_{A}: \mu>\mu_{0}, n=28, T=2.13\)

Short Answer

Expert verified
Reject null hypothesis for (a), (b), and (d); do not reject for (c).

Step by step solution

01

Identify the Test Type and Tail Direction

The hypotheses are given as directional: (a) is a one-tailed test (right tail) because \(H_{A}: \mu > \mu_{0}\).(b) is a one-tailed test (left tail) because \(H_{A}: \mu < \mu_{0}\).(c) is a two-tailed test since \(H_{A}: \mu eq \mu_{0}\).(d) is a one-tailed test (right tail) because \(H_{A}: \mu > \mu_{0}\).
02

Determine the Degrees of Freedom

The degrees of freedom (\(df\)) are calculated as \(df = n - 1\).(a) \(df = 11 - 1 = 10\)(b) \(df = 17 - 1 = 16\)(c) \(df = 7 - 1 = 6\)(d) \(df = 28 - 1 = 27\)
03

Look up the p-value for each T statistic

Using a T-table or calculator, find the p-values for the given T statistic and degrees of freedom.(a) \(T = 1.91, df = 10\): \(p \approx 0.043\) (right-tailed)(b) \(T = -3.45, df = 16\): \(p \approx 0.001 \) (left-tailed)(c) \(T = 0.83, df = 6\): \(p \approx 0.441 \) (two-tailed)(d) \(T = 2.13, df = 27\): \(p \approx 0.021 \) (right-tailed)
04

Compare p-value to significance level \(\alpha=0.05\)

Check each p-value against \(\alpha = 0.05\): (a) \(p = 0.043 < 0.05\), reject the null hypothesis.(b) \(p = 0.001 < 0.05\), reject the null hypothesis.(c) \(p = 0.441 > 0.05\), do not reject the null hypothesis.(d) \(p = 0.021 < 0.05\), reject the null hypothesis.

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Key Concepts

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

t-distribution
The t-distribution is a fundamental concept in statistics used when dealing with small sample sizes and an unknown population standard deviation. Unlike the normal distribution, the t-distribution accounts for the additional uncertainty that accompanies small sample sizes.
The shape of the t-distribution is similar to the normal distribution but has thicker tails, which means it is more prone to extreme values. These thicker tails ensure that the t-distribution provides more conservative estimates, which is crucial for hypothesis testing.
  • The distribution is symmetric around zero, just like a normal distribution.
  • As the sample size increases, the t-distribution approaches the normal distribution.
  • It is indexed by degrees of freedom, which affect the shape of the distribution.
Understanding the t-distribution is crucial when conducting hypothesis tests involving the t-statistic.
hypothesis testing
Hypothesis testing is a statistical method used to make inferences about population parameters based on sample data. The process begins with formulating two opposing hypotheses: the null hypothesis ( H_0 ) and the alternative hypothesis ( H_A ).
The null hypothesis typically represents the status quo or a statement of no effect, while the alternative hypothesis indicates the presence of an effect or difference.

After collecting data, a test statistic is computed and compared against a critical value, determined by the significance level and the distribution of the test statistic.
  • If the test statistic falls within the rejection region, the null hypothesis is rejected in favor of the alternative hypothesis.
  • If it falls outside, the null hypothesis is not rejected.
One important part of hypothesis testing is understanding whether the test is one-tailed or two-tailed, as this affects the interpretation of the test statistic and p-value.
degrees of freedom
Degrees of freedom (df) are a key concept in statistics and play a major role in determining the shape of the t-distribution used in hypothesis testing. Simply put, degrees of freedom are the number of independent values or quantities that can freely vary in a statistical calculation.

In the context of a t-test, degrees of freedom are calculated as the sample size minus one ( {df = n - 1} ). This calculation is vital because it influences the critical value you look up when using t-tables to find p-values.
  • A smaller number of degrees of freedom typically results in a t-distribution with wider tails.
  • Larger degrees of freedom make the t-distribution more closely resemble a normal distribution, smoothing out the tails.
Grasping the concept of degrees of freedom is crucial for accurate hypothesis testing, as it directly affects the outcome and interpretation of your results.
significance level
The significance level, often denoted by \(\alpha\), is the threshold used in hypothesis testing to determine whether to reject the null hypothesis. It represents the probability of incorrectly rejecting the null hypothesis when it is true (Type I error).

In practice, common significance levels are 0.05, 0.01, and 0.10, with 0.05 being the most widely used. The chosen significance level impacts the critical region, or the range of values in which the null hypothesis would be rejected.
  • A lower significance level means a smaller critical region, increasing the stringency of the test.
  • Conversely, a higher significance level allows for a broader critical region and a higher chance of rejecting the null hypothesis.
Recognizing how the significance level influences hypothesis testing is key to making informed decisions from statistical findings.

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Most popular questions from this chapter

Diamonds, Part I. Prices of diamonds are determined by what is known as the 4 Cs: cut, clarity, color, and carat weight. The prices of diamonds go up as the carat weight increases, but the increase is not smooth. For example, the difference between the size of a 0.99 carat diamond and a 1 carat diamond is undetectable to the naked human eye, but the price of a 1 carat diamond tends to be much higher than the price of a 0.99 diamond. In this question we use two random samples of diamonds, 0.99 carats and 1 carat, each sample of size \(23,\) and compare the average prices of the diamonds. In order to be able to compare equivalent units, we first divide the price for each diamond by 100 times its weight in carats. That is, for a 0.99 carat diamond, we divide the price by 99. For a 1 carat diamond, we divide the price by \(100 .\) The distributions and some sample statistics are shown below. \(^{37}\) Conduct a hypothesis test to evaluate if there is a difference between the average standardized prices of 0.99 and 1 carat diamonds. Make sure to state your hypotheses \begin{tabular}{|l|} \hline \\ \hline \end{tabular} clearly, check relevant conditions, and interpret your results in context of the data.

True or false, Part II. Determine if the following statements are true or false in ANOVA, and explain your reasoning for statements you identify as false. (a) As the number of groups increases, the modified significance level for pairwise tests increases as well. (b) As the total sample size increases, the degrees of freedom for the residuals increases as well. (c) The constant variance condition can be somewhat relaxed when the sample sizes are relatively consistent across groups. (d) The independence assumption can be relaxed when the total sample size is large.

Coffee, depression, and physical activity. Caffeine is the world's most widely used stimulant, with approximately \(80 \%\) consumed in the form of coffee. Participants in a study investigating the relationship between coffee consumption and exercise were asked to report the number of hours they spent per week on moderate (e.g., brisk walking) and vigorous (e.g., strenuous sports and jogging) exercise. Based on these data the researchers estimated the total hours of metabolic equivalent tasks (MET) per week, a value always greater than \(0 .\) The table below gives summary statistics of MET for women in this study based on the amount of coffee consumed. 3 (a) Write the hypotheses for evaluating if the average physical activity level varies among the different levels of coffee consumption. (b) Check conditions and describe any assumptions you must make to proceed with the test. (c) Below is part of the output associated with this test. Fill in the empty cells. (d) What is the conclusion of the test?

4.36 True or false, Part I. Determine if the following statements are true or false, and explain your reasoning for statements you identify as false. (a) When comparing means of two samples where \(n_{1}=20\) and \(n_{2}=40,\) we can use the normal model for the difference in means since \(n_{2} \geq 30\). (b) As the degrees of freedom increases, the T distribution approaches normality. (c) We use a pooled standard error for calculating the standard error of the difference between means when sample sizes of groups are equal to each other.

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