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Chapter 4: Problem 1

Identify the critical t. An independent random sample is selected from an approximately normal population with unknown standard deviation. Find the degrees of freedom and the critical \(t\) value \(\left(\mathrm{t}^{*}\right)\) for the given sample size and confidence level. (a) \(n=6, \mathrm{CL}=90 \%\) (c) \(n=29, \mathrm{CL}=95 \%\) (b) \(n=21, \mathrm{CL}=98 \%\) (d) \(n=12, \mathrm{CL}=99 \%\)

Short Answer

Expert verified
(a) df = 5, t* 鈮� 2.015; (b) df = 20, t* 鈮� 2.528; (c) df = 28, t* 鈮� 2.048; (d) df = 11, t* 鈮� 3.106.

Step by step solution

01

Calculating Degrees of Freedom for Part (a)

First, determine the degrees of freedom (df) for part (a) with sample size \( n = 6 \). The formula is \( df = n - 1 \). Thus, for this sample size, \( df = 6 - 1 = 5 \).
02

Finding Critical t Value for Part (a)

With \( df = 5 \) and a confidence level of 90%, look up the critical t value (\( t^* \)) from the t-distribution table. The critical t value for these parameters is approximately \( t^* = 2.015 \).
03

Calculating Degrees of Freedom for Part (b)

Determine the degrees of freedom for part (b) with \( n = 21 \). Use the formula \( df = n - 1 \) resulting in \( df = 21 - 1 = 20 \).
04

Finding Critical t Value for Part (b)

With \( df = 20 \) and a confidence level of 98%, the critical t value (\( t^* \)) from the t-distribution table is approximately \( t^* = 2.528 \).
05

Calculating Degrees of Freedom for Part (c)

Calculate the degrees of freedom for part (c) with \( n = 29 \). The formula \( df = n - 1 \) gives \( df = 29 - 1 = 28 \).
06

Finding Critical t Value for Part (c)

With \( df = 28 \) and a confidence level of 95%, the critical t value from the t-distribution table is \( t^* = 2.048 \).
07

Calculating Degrees of Freedom for Part (d)

Now for part (d) with \( n = 12 \), compute the degrees of freedom as \( df = 12 - 1 = 11 \).
08

Finding Critical t Value for Part (d)

With \( df = 11 \) and confidence level 99%, refer to the t-distribution table to find that \( t^* = 3.106 \).

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Key Concepts

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

Degrees of Freedom
Degrees of freedom are used in statistics to describe the number of values in a calculation that are free to vary. When you have a sample of data and you're estimating population parameters, degrees of freedom help in assessing how much information is really available to make inferences about the population. For the t-distribution, which is used when dealing with small sample sizes or unknown population standard deviations, the formula for calculating the degrees of freedom is quite straightforward: subtract 1 from the sample size.
  • For a sample size of 6, the degrees of freedom is calculated as 6 - 1 = 5.
  • If you have a sample size of 21, it would be 21 - 1 = 20.
  • A sample size of 29 leads to 29 - 1 = 28 degrees of freedom.
  • For a sample of 12, the degrees of freedom is 12 - 1 = 11.
This simple concept is the first step in interpreting statistical results that involve the t-distribution.
Critical t Value
The critical t value is a threshold value that dictates whether the null hypothesis can be rejected in hypothesis testing. It depends on both the confidence level and the degrees of freedom. The t value indicates how extreme an observed statistic must be for it to be considered significantly different from what was expected under the null hypothesis. To find the critical t value for a specific test:
  • Identify the degrees of freedom, which comes from the sample size (minus one).
  • Determine the desired confidence level for the test.
  • Use a t-distribution table or statistical software to find the critical value that corresponds to those degrees of freedom and confidence level.
For example, with 5 degrees of freedom and a 90% confidence level, the critical t value is approximately 2.015. Adjusting the degrees of freedom or confidence level will change the critical t value, illustrating its dependency on both parameters.
Confidence Level
The confidence level represents the degree of certainty that the parameter falls within the range of values defined by the statistical test. It is expressed as a percentage and conveys how confident you can be that the interval estimate contains the population parameter. Common confidence levels include:
  • 90% confidence level, offering a moderate level of confidence that the true parameter lies within the interval.
  • 95% confidence level, which is widely used and provides a higher degree of confidence compared to 90%.
  • 98% and 99% confidence levels, offering very high confidence at the cost of wider intervals.
In the given exercise, different confidence levels are paired with varying sample sizes. This affects the critical t value used in determining the interval, as higher confidence levels require higher critical t values.
Sample Size
Sample size refers to the number of observations included in a sample. It plays an essential role in statistical inference because it affects the degrees of freedom and consequently the critical t value. Here are some impacts and considerations of sample size:
  • Smaller sample sizes typically mean less accurate data representation of a population, leading to a higher critical t value for a given confidence level.
  • Larger sample sizes tend to yield more reliable estimates, reducing the critical t value needed for the same confidence level.
  • As sample sizes increase, results more closely approximate a normal distribution.
In any statistical analysis using t-distributions, understanding how your sample size affects your calculations is crucial. This can lead to more informed conclusions about the data at hand. Optimizing sample size helps in arriving at precise and meaningful statistical inferences.

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Most popular questions from this chapter

Body fat in women and men. The third National Health and Nutrition Examination Survey collected body fat percentage (BF) data from 13,601 subjects whose ages are 20 to \(80 .\) A summary table for these data is given below. Note that BF is given as mean \(\pm\) standard error. Construct a \(95 \%\) confidence interval for the difference in average body fat percentages between men and women, and explain the meaning of this interval. Tip: the standard error can be calculated as \(S E=\sqrt{S E_{M}^{2}+S E_{W}^{2}}\)

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.

Does the Paleo diet work? The Paleo diet allows only for foods that humans typically consumed over the last 2.5 million years, excluding those agriculture-type foods that arose during the last 10,000 years or so. Researchers randomly divided 500 volunteers into two equal-sized groups. One group spent 6 months on the Paleo diet. The other group received a pamphlet about controlling portion sizes. Randomized treatment assignment was performed, and at the beginning of the study, the average difference in weights between the two groups was about 0. After the study, the Paleo group had lost on average 7 pounds with a standard deviation of 20 pounds while the control group had lost on average 5 pounds with a standard deviation of 12 pounds. (a) The \(95 \%\) confidence interval for the difference between the two population parameters (Paleo \- control) is given as (-0.891,4.891) . Interpret this interval in the context of the data. (b) Based on this confidence interval, do the data provide convincing evidence that the Paleo diet is more effective for weight loss than the pamphlet (control)? Explain your reasoning. (c) Without explicitly performing the hypothesis test, do you think that if the Paleo group had lost 8 instead of 7 pounds on average, and everything else was the same, the results would then indicate a significant difference between the treatment and control groups? Explain your reasoning.

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) \(H_{A}: \mu>\mu_{0}, n=28, T=2.13\)

True or false, Part III. Determine if the following statements are true or false, and explain your reasoning for statements you identify as false. If the null hypothesis that the means of four groups are all the same is rejected using ANOVA at a \(5 \%\) significance level, then ... (a) we can then conclude that all the means are different from one another. (b) the standardized variability between groups is higher than the standardized variability within groups. (c) the pairwise analysis will identify at least one pair of means that are significantly different. (d) the appropriate \(\alpha\) to be used in pairwise comparisons is \(0.05 / 4=0.0125\) since there are four groups.
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