Use the listed paired sample data, and assume that the...

Use the listed paired sample data, and assume that the samples are simple random samples and that the differences have a distribution that is approximately normal. Listed below are body temperatures from seven different subjects measured at two different times in a day (from Data Set 3 "Body Temperatures" in Appendix B). a. Use a 0.05 significance level to test the claim that there is no difference between body temperatures measured at $8 \mathrm{AM}$ and at $12 \mathrm{AM}$. b. Construct the confidence interval that could be used for the hypothesis test described in part (a). What feature of the confidence interval leads to the same conclusion reached in part (a)? $$\begin{array}{l|l|l|l|l|l|l|l} \hline \text { Body Temperature ("F) at 8 AM } & 96.6 & 97.0 & 97.0 & 97.8 & 97.0 & 97.4 & 96.6 \\ \hline \text { Body Temperature ('F) at 12 AM } & 99.0 & 98.4 & 98.0 & 98.6 & 98.5 & 98.9 & 98.4 \\ \hline \end{array}$$

Use the listed paired sample data, and assume that the samples are simple random samples and that the differences have a distribution that is approximately normal.
Listed below are body temperatures from seven different subjects measured at two different times in a day (from Data Set 3 "Body Temperatures" in Appendix B).
a. Use a 0.05 significance level to test the claim that there is no difference between body temperatures measured at $8 \mathrm{AM}$ and at $12 \mathrm{AM}$.
b. Construct the confidence interval that could be used for the hypothesis test described in part
(a). What feature of the confidence interval leads to the same conclusion reached in part (a)?
$$\begin{array}{l|l|l|l|l|l|l|l}
\hline \text { Body Temperature ("F) at 8 AM } & 96.6 & 97.0 & 97.0 & 97.8 & 97.0 & 97.4 & 96.6 \\
\hline \text { Body Temperature ('F) at 12 AM } & 99.0 & 98.4 & 98.0 & 98.6 & 98.5 & 98.9 & 98.4 \\
\hline
\end{array}$$

Key Concepts

Significance Level

The significance level, often denoted by alpha, is the criterion used to decide whether the results of a hypothesis test are statistically significant. A commonly used significance level, such as 0.05, indicates that there is a 5% risk of rejecting the null hypothesis when it is actually true. This threshold guides the decision-making process in hypothesis testing.

Confidence Interval

A confidence interval is a range of values derived from sample data that is likely to contain the true population parameter. For paired data, the confidence interval for the mean difference provides insight into the magnitude and precision of the effect. If this interval does not include the value specified under the null hypothesis (often zero), it leads to rejecting the null hypothesis.

Hypothesis Testing

Hypothesis testing is a process of making statistical decisions using experimental data. In the context of comparing paired measurements, the null hypothesis typically states that there is no difference between the two conditions, while the alternative hypothesis asserts that a difference exists. The test uses a significance level to decide whether to reject the null hypothesis.

Paired t-Test

A paired t-test is a statistical method used to determine whether the mean difference between paired observations is zero. It is specifically used when the differences between paired measurements are assumed to be approximately normally distributed, and it accounts for the fact that the observations are not independent.

Paired Sample Data

Paired sample data involve measurements taken on the same subjects or matched subjects under two different conditions. This approach reduces variability by using each subject as their own control, making it easier to detect differences between the conditions being compared.

Transcript

00:01 So for this question, i'm going to have my difference be the 8 a .m.

00:04 Temp minus the 12 a .m.

00:08 Temp.

00:09 And i have my differences as negative 2 .4, negative 1 .4, negative 1 .8, negative 1 .5, negative 1 .5, and negative 1 .8.

00:22 And so i have a total of seven pieces of data.

00:25 I have that the mean of those differences is equal to negative 1 .4857.

00:31 And i have the standard deviation of those difference is equal to 0 .5242.

00:37 And so we will be assuming that the mean difference is actually equal to zero.

00:44 And alternately, that it's different.

00:47 Oops, mean difference is different from zero.

00:50 So we're doing a two -tail test.

00:53 And we want to find in the picture, we are again assuming that that mean difference is at zero, but we're getting a value of negative, 1 .4857.

01:07 And since we're doing a two -tail test, we're going to, that shaded area, oh, and i shade it in red, that shaded area is going to be my p value.

01:15 So let's continue on with our test.

01:18 So we want to find what's the likelihood of sampling and getting a test statistic that is, and let's see, well, let me write it up first, a d bar that is less than or equal to negative 1 .4857.

01:36 So that's a d bar.

01:38 We want to find what this likelihood is and double it...

Please give Ace some feedback