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\(t\) -models, part I Using the \(t\) tables, software, or a calculator, estimate a) the critical value of \(t\) for a \(90 \%\) confidence interval with df \(=17\) b) the critical value of \(t\) for a \(98 \%\) confidence interval with df \(=88\) c) the P-value for \(t \geq 2.09\) with 4 degrees of freedom. d) the P-value for \(|t|>1.78\) with 22 degrees of freedom.

Short Answer

Expert verified
a) Critical value ≈ 1.740; b) Critical value ≈ 2.637; c) P-value ≈ 0.055; d) P-value ≈ 0.088.

Step by step solution

01

Determine the Critical Value for df = 17, 90% Confidence Interval

To find the critical value of the t-distribution at a 90% confidence interval with 17 degrees of freedom, we need to find the t-score that captures the middle 90% of the distribution. This implies we leave out 5% on each tail since it is two-tailed, producing a total missing 10% cumulatively. Using a t-table or calculator, we look up the row for 17 degrees of freedom and locate the value at 0.05 probability on the one tail (since it is a 90% confidence level, which corresponds to 0.10 in two tails). The critical value is approximately 1.740.
02

Determine the Critical Value for df = 88, 98% Confidence Interval

For a 98% confidence interval with 88 degrees of freedom, we aim for a span that captures 98% of the distribution's probability, which leaves 1% on each tail in a two-tailed test. Checking the t-table or using a calculator for 88 degrees of freedom and a probability level of 0.01 on one tail (because 98% CI implies 0.02 split across both tails), the critical value approximately is 2.637.
03

Calculate the P-value for t ≥ 2.09, df = 4

When calculating the p-value for t ≥ 2.09 with 4 degrees of freedom, we're looking at the area in the upper tail at t = 2.09. Using a t-distribution table or a calculator, we find the area to the right of t = 2.09. For 4 degrees of freedom, p(t ≥ 2.09) is approximately 0.055. Thus, the p-value is around 0.055.
04

Determine the P-value for |t| > 1.78, df = 22

To find the p-value for |t| > 1.78 with 22 degrees of freedom, we calculate the probability of t being less than -1.78 or greater than 1.78, as it's a two-tailed query. Using a table or software, we determine the area in both tails beyond ±1.78. For 22 degrees of freedom, p(|t| > 1.78) corresponds to roughly 0.088. Hence, the p-value is about 0.088.

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Key Concepts

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

Confidence Interval
A confidence interval provides us with a range where we expect a population parameter to fall with a certain level of certainty. For example, in the exercise, we determined the critical value for confidence intervals of 90% and 98%.
For a 90% confidence interval, we are saying that if we took many samples and calculated the interval for each, we expect 90% of those intervals to contain the true population parameter.
This involves leaving out 5% from each tail of the t-distribution because the entire 10% remains outside our interval.

Confidence intervals help assess the reliability of an estimate. The larger the confidence level, such as 98%, the wider the interval, which reflects greater certainty. This is crucial even if it sometimes results in less precision in practical applications.
Degrees of Freedom
Degrees of freedom (df) in statistics resemble the number of values in a calculation that are free to vary. When estimating a parameter, especially in tests involving the t-distribution, knowing the degrees of freedom is essential.
In the t-distribution tables, df influences the shape of the curve. Lower degrees of freedom result in a curve that's more spread out or 'flatter,' indicating more variability.
This relates to the exercise, such as 17 and 88 df for our confidence intervals, and 4 and 22 df for determining the P-value.

As the degrees of freedom increase (generally aligning with larger sample sizes), the t-distribution resembles the normal distribution more closely, making it crucial in determining critical values and P-values.
Critical Value
A critical value is a point on a statistical distribution that marks an area where the null hypothesis will be rejected. In relation to a confidence interval, it defines the boundary values of the interval.
For example, in the exercise, we found critical values like 1.740 for a 90% confidence interval with 17 df, and 2.637 for a 98% confidence interval with 88 df.

These values effectively determine the 'cut-off' points of the tails of the distribution beyond which we consider results statistically significant. Critical values are determined from the t-distribution table based on confidence levels and degrees of freedom. They help guide conclusions about the data from samples regarding hypotheses or interval estimates.
P-value
A P-value is a metric that helps us understand how extreme observed data is under the null hypothesis. It gives a probability measure: the probability of seeing the observed data, or something more extreme, given that the null hypothesis is true.
Looking at the exercise, we calculated P-values for specific t-values beyond certain points. For instance, for t ≥ 2.09 with df = 4, the P-value was about 0.055, and for |t| > 1.78 with df = 22, it was approximately 0.088.

A low P-value (usually less than 0.05) indicates strong evidence against the null hypothesis, implying the observed effect is statistically significant. Understanding P-values is essential in gauging the strength of evidence and making data-driven decisions.

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Most popular questions from this chapter

LSAT The LSAT (a test taken for law school admission) has a mean score of 151 with a standard deviation of 9 and a unimodal, symmetric distribution of scores. A test preparation organization teaches small classes of 9 students at a time. A larger organization teaches classes of 25 students at a time. Both organizations publish the mean scores of all their classes. a) What would you expect the distribution of mean class scores to be for each organization? b) If cither organization has a graduating class with a mean score of \(160,\) they'll take out a full-page ad in the local school paper to advertise. Which organization is more likely to have that success? Explain. c) Both organizations advertise that if any class has an average score below \(145,\) they'll pay for everyone to retake the LSAT. Which organization is at greater risk to have to pay?

Hot Dogs A nutrition lab tested 40 hot dogs to see if their mean sodium content was less than the 325 mg upper limit set by regulations for "reduced sodium" franks. The lab failed to reject the hypothesis that the hot dogs did not meet this requirement, with a P-value of \(0.142 . \mathrm{A} 90 \%\) confidence interval estimated the mean sodium content for this kind of hot dog at 317.2 to 326.8 mg. Explain how these two results are consistent. Your explanation should discuss the confidence level, the P-value, and the decision.

Chips Ahoy In \(1998,\) as an advertising campaign, the Nabisco Company announced a " 1000 Chips Challenge," claiming that every 18 -ounce bag of their Chips Ahoy cookies contained at least 1000 chocolate chips. Dedicated Statistics students at the Air Force Academy (no kidding) purchased some randomly selected bags of cookies, and counted the chocolate chips. Some of their data are given below. (Chance, \(12,\) no. 1[1999] ) $$\begin{array}{llllllll} 1219 & 1214 & 1087 & 1200 & 1419 & 1121 & 1325 & 1345 \\ 1244 & 1258 & 1356 & 1132 & 1191 & 1270 & 1295 & 1135 \end{array}$$ a) Check the assumptions and conditions for inference. Comment on any concerns you have. b) Create a \(95 \%\) confidence interval for the average number of chips in bags of Chips Ahoy cookies. c) What does this evidence say about Nabisco's claim? Use your confidence interval to test an appropriate hypothesis and state your conclusion.

Popcorn Yvon Hopps ran an experiment to test optimum power and time settings for microwave popcorn. His goal was to find a combination of power and time that would deliver high-quality popcorn with less than \(10 \%\) of the kernels left unpopped, on average. After experimenting with several bags, he determined that power 9 at 4 minutes was the best combination. a) He concluded that this popping method achieved the \(10 \%\) goal. If it really does not work that well, what kind of error did Hopps make? b) To be sure that the method was successful, he popped 8 more bags of popcorn (selected at random) at this setting. All were of high quality, with the following percentages of uncooked popcorn: 7,13.2,10,6,7.8 \(2.8,2.2,5.2 .\) Does this provide evidence that he met his goal of an average of no more than \(10 \%\) uncooked kernels? Explain.

\(t\) -models, part II Using the \(t\) tables, software, or a calculator, estimate a) the critical value of \(t\) for a \(95 \%\) confidence interval with df \(=7\) b) the critical value of \(t\) for a \(99 \%\) confidence interval with \(\mathrm{df}=102\) c) the P-value for \(t \leq 2.19\) with 41 degrees of freedom. d) the P-value for \(|t|>2.33\) with 12 degrees of freedom.

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