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Find the tabled value of \(t\left(t_{a}\right)\) corresponding to a left-tail area a and degrees of freedom given. $$ a=.10, d f=35 $$

Short Answer

Expert verified
Answer: To find the tabled value of the t-distribution with a left-tail area of 0.10 and degrees of freedom 35, follow these steps: 1. Locate the row corresponding to 35 degrees of freedom in a t-distribution table. 2. Locate the column corresponding to the left-tail area of 0.10 in the t-distribution table. 3. Identify the value at the intersection of the row and column located in Steps 1 and 2. 4. The tabled value of \(t\left(t_{a}\right)\) at the intersection is the critical value of the t-distribution for this problem. Consult a physical or online t-distribution table to find the exact value.

Step by step solution

01

Locate the degrees of freedom (df) in the t-distribution table.

In the t-distribution table, locate the row corresponding to the given degrees of freedom (df), which is 35 in this case.
02

Locate the left-tail area (a) in the t-distribution table.

In the t-distribution table, locate the column corresponding to the given left-tail area (a), which is 0.10 in this case.
03

Find the intersection of the corresponding row and column.

Identify the value at the intersection of the row and column located in Steps 1 and 2. This value represents the critical value of the t-distribution for the given left-tail area and degrees of freedom.
04

Write down the tabled value of \(t\left(t_{a}\right)\).

The tabled value of \(t\left(t_{a}\right)\) at the intersection of the row with degrees of freedom 35 and column with left-tail area 0.10 is the critical value of the t-distribution for this problem. Note: Since we don't have a specific t-distribution table to look at, you'll need to consult a physical or online t-distribution table to find the exact value for this problem. These steps can be followed for any t-distribution table.

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Key Concepts

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

Degrees of Freedom
Understanding the concept of 'degrees of freedom' (df) is crucial when working with statistics, especially in hypothesis testing and estimation problems. In simple terms, degrees of freedom refer to the number of values in a statistical calculation that are free to vary. To illustrate, imagine you have a dataset of five test scores and you know their average. If you have four of the tests scores, you can calculate the fifth one because it is not 'free'—it is constrained by the known average.
  • When dealing with the t-distribution, degrees of freedom are determined by the sample size minus one (-df = n-1-); where -n- is the number of observations or sample size.
  • The degrees of freedom are used to determine the shape of the t-distribution curve. As the degrees of freedom increase, the t-distribution approaches a normal distribution.
  • In practical terms, higher degrees of freedom generally result in a t-curve that is flatter and more spread out. This means that the critical values will get closer to those of the normal distribution as the degrees of freedom increase.

The exercise provided asks for a calculation using 35 degrees of freedom. As the given df is relatively high, the shape of the t-distribution would be expected to closely resemble a standard normal distribution, though still with slightly heavier tails.
Left-tail Area
The 'left-tail area' in a t-distribution table represents the probability that a t-value will fall to the left of a specified value. It is a critical concept when performing left-tailed hypothesis tests, which typically involve testing whether a mean is less than a certain value.
  • The left-tail area is denoted by 'a' and in our exercise, it is 0.10. It is often associated with the significance level of a test. In our case, having a left-tail area of 0.10 means that there is a 10% probability that the t-value will be less than the critical value.
  • This area is important because it helps us to find the critical value of t, which determines the threshold for rejecting the null hypothesis. If the calculated test statistic is less than this critical value, the null hypothesis is rejected in favor of the alternative hypothesis.

Locating the left-tail area in the t-distribution table is as simple as finding the appropriate column titled with the desired probability, which corresponds directly to the level of significance for the test.
Critical Value of t-distribution
The 'critical value of t-distribution' is the t-value that separates the region where the null hypothesis is not rejected from where it is rejected. This value is found in the t-distribution table, given a specific left-tail area (or significance level) and degrees of freedom associated with the statistical test.
  • To locate the critical value, one must intersect the row corresponding to the degrees of freedom with the column for the desired left-tail area in the t-distribution table.
  • The critical value represents a threshold: if the calculated t-value from a test statistic falls beyond this value in the left tail, we have sufficient evidence to reject the null hypothesis at the given significance level.
  • The closer the left-tail area is to 0.5, the smaller the absolute value of the critical value, indicating a larger area where the null hypothesis would be rejected.

For the exercise problem, the critical t-value at 35 degrees of freedom and a left-tail area of 0.10 would be found at the intersection given in the t-distribution table. This is the value required to determine the outcome of the statistical test based on the t-distribution.

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Most popular questions from this chapter

The cost of auto insurance in California is dependent on many variables, such as the city you live in, the number of cars you insure, and your insurance company. The website www.insurance.ca.gov reports the annual 2017 standard premium for a male, licensed for \(6-8\) years, who drives a Honda Accord 20,000 to 24,000 kilometers per year and has no violations or accidents. \({ }^{12}\) $$\begin{array}{lll}\hline \text { City } & \text { Allstate } & \text { 21st Century } \\\\\hline \text { Long Beach } & \$ 3447 & \$ 3156 \\\\\text { Pomona } & 3572 & 3108 \\\\\text { San Bernardino } & 3393 & 3110 \\\\\text { Moreno Valley } & 3492 & 3300 \\\\\hline\end{array}$$ a. Why would you expect these pairs of observations to be dependent? b. Do the data provide sufficient evidence to indicate that there is a difference in the average annual premiums between Allstate and 21 st Century insurance? Test using \(\alpha=.01\) c. Find the approximate \(p\) -value for the test and interpret its value. d. Find a \(99 \%\) confidence interval for the difference in the average annual premiums for Allstate and 21 st Century insurance. e. Can we use the information in the table to make valid comparisons between Allstate and 21 st Century insurance throughout the United States? Why or why not?

A pharmaceutical manufacturer is concerned about the variability of the impurities from shipment to shipment from two different suppliers. To compare the variation in percentage impurities, the manufacturer selects 10 shipments from each of the two suppliers and measures the percentage of impurities in the raw material for each shipment. The sample means and variances are shown in the table. $$\begin{array}{ll}\hline \text { Supplier A } & \text { Supplier B } \\\\\hline \bar{x}_{1}=1.89 & \bar{x}_{2}=1.85 \\\s_{1}^{2}=.273 & s_{2}^{2}=.094 \\\n_{1}=10 & n_{2}=10\end{array}$$ a. Do the data provide sufficient evidence to indicate a difference in the variability of the shipment impurity levels for the two suppliers? Test using \(\alpha=.01 .\) Based on the results of your test, what recommendation would you make to the pharmaceutical manufacturer? b. Find a \(99 \%\) confidence interval for \(\sigma_{2}^{2}\) and interpret your results.

To compare the demand for two different entrees, \(A\) and \(B\), a cafeteria manager recorded the number of purchases of each entree on seven consecutive days. Do the data provide sufficient evidence to indicate a greater mean demand for one of the entrees? Use the Excel printout. $$\begin{array}{lcc}\hline \text { Day } & \mathrm{A} & \mathrm{B} \\\\\hline \text { Monday } & 420 & 391 \\\\\text { Tuesday } & 374 & 343 \\\\\text { Wednesday } & 434 & 469 \\\\\text { Thursday } & 395 & 412 \\\\\text { Friday } & 637 & 538 \\\\\text { Saturday } & 594 & 521 \\\\\text { Sunday } & 679 & 625 \\\\\hline\end{array}$$

A production plant has two fabricating systems, both of which are maintained at 2-week intervals. However, one system is twice as old as the other. The number of finished products fabricated daily by each of the systems is recorded for 30 working days, with the results given in the table. Do these data present sufficient evidence to conclude that the variability in daily production warrants increased maintenance of the older fabricating system? Use the \(p\) -value approach. $$\begin{array}{ll}\hline \text { New System } & \text { Old System } \\\\\hline \bar{x}_{1}=246 & \bar{x}_{2}=240 \\\s_{1}=15.6 & s_{2}=28.2 \\\\\hline\end{array}$$

To compare the mean lengths of time required for the bodily absorption of two drugs A and \(\mathrm{B}, 10\) people were randomly selected and assigned to receive one of the drugs. The length of time (in minutes) for the drug to reach a specified level in the blood was recorded, and the data summary is given in the table: $$ \begin{array}{ll} \hline \text { Drug A } & \text { Drug B } \\ \hline \bar{x}_{1}=27.2 & \bar{x}_{2}=33.5 \\ s_{1}^{2}=16.36 & s_{2}^{2}=18.92 \\ \hline \end{array} $$ a. Do the data provide sufficient evidence to indicate a difference in mean times to absorption for the two drugs? Test using \(\alpha=.05 .\) b. Find the approximate \(p\) -value for the test. Does this value confirm your conclusions? c. Find a \(95 \%\) confidence interval for the difference in mean times to absorption. Does the interval confirm your conclusions in part a?

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