respectively. Assume that the bond strength distributions are both normal. (a) Assuming that \( \sigma_{1}=1.6 \) and \( \sigma_{2}=1.3 \), test \( H_{0}: \mu_{1}-\mu_{2}=0 \) versus \( H_{a}: \mu_{1}-\mu_{2}>0 \) at level 0.01 . Calculate the test statistic and determine the \( P \)-value. (Round your test statistic to two decimal places and your \( P \)-value to four decimal places.) \( \begin{aligned} z & = \\ P \text {-value } & =\end{aligned} \) State the conclusion in the problem context. Fail to reject \( H_{0} \). The data suggests that the difference in average tension bond strengths exceeds 0 . Fail to reject \( H_{0} \). The data does not suggest that the difference in average tension bond strengths exceeds from 0 . Reject \( H_{0} \). The data suggests that the difference in average tension bond strengths exceeds 0 . Reject \( H_{0} \). The data does not suggest that the difference in average tension bond strengths exceeds 0 . (b) Compute the probability of a type II error for the test of part (a) when \( \mu_{1}-\mu_{2}=1 \). (Round your answer to four decimal places.) \( n= \) (d) How would the analysis and conclusion of part (a) change if \( \sigma_{1} \) and \( \sigma_{2} \) were unknown but \( s_{1}=1.6 \) and \( s_{2}=1.3 \) ? Since \( n=32 \) is not a large sample, it still be appropriate to use the large sample test. The analysis and conclusions would stay the same. Since \( n=32 \) is a large sample, it would be more appropriate to use the \( t \) procedure. The appropriate conclusion would follow. Since \( n=32 \) is a large sample, it would no longer be appropriate to use the large sample test. Any other test can be used, and the conclusions would stay the same.
Added by Unknown S.
Close
Step 1
(a) Show more…
Show all steps
Your feedback will help us improve your experience
Megha Sharma and 101 other Intro Stats / AP Statistics educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
I need help with this
Supreeta N.
What about the sample size $n$ for confidence intervals for the difference of proportions $p_{1}-p_{2} ?$ Let us make the following assumptions: equal sample sizes $n=n_{1}=n_{2}$ and all four quantities $n_{1} \hat{p}_{1}, n_{1} \hat{q}_{1}, n_{2} \hat{p}_{2},$ and $n_{2} \hat{q}_{2}$ are greater than $5 .$ Those readers familiar with algebra can use the procedure outlined in Problem 28 to show that if we have preliminary estimates $\hat{p}_{1}$ and $\hat{p}_{2}$ and a given maximal margin of error $E$ for a specified confidence level $c,$ then the sample size $n$ should be at least $$n=\left(\frac{z_{c}}{E}\right)^{2}\left(\hat{p}_{1} \hat{q}_{1}+\hat{p}_{2} \hat{q}_{2}\right)$$ However, if we have no preliminary estimates for $\hat{p}_{1}$ and $\hat{p}_{2}$, then the theory similar to that used in this section tells us that the sample size $n$ should be at least $$n=\frac{1}{2}\left(\frac{z_{c}}{F}\right)^{2}$$ (a) In Problem 17 (Myers-Briggs personality type indicators in common for married couples), suppose we want to be $99 \%$ confident that our estimate $\hat{p}_{1}-\hat{p}_{2}$ for the difference $p_{1}-p_{2}$ has a maximal margin of error $E=0.04 .$ Use the preliminary estimates $\hat{p}_{1}=289 / 375$ for the proportion of couples sharing two personality traits and $\hat{p}_{2}=23 / 571$ for the proportion having no traits in common. How large should the sample size be (assuming equal sample size-i.e., $n=n_{1}=n_{2}$ )? (b) Suppose that in Problem 17 we have no preliminary estimates for $\hat{p}_{1}$ and $\hat{p}_{2}$ and we want to be $95 \%$ confident that our estimate $\hat{p}_{1}-\hat{p}_{2}$ for the difference $p_{1}-p_{2}$ has a maximal margin of error $E=0.05 .$ How large should the sample size be (assuming equal sample size-i.e., $n=n_{1}=n_{2}$ )?
Estimation
Estimating $\mu_{1}-\mu_{2}$ and $p_{1}-p_{2}$
Is there a relationship between confidence intervals and two-tailed hypothesis tests? Let $c$ be the level of confidence used to construct a confidence interval from sample data. Let $\alpha$ be the level of significance for a two-tailed hypothesis test. The following statement applies to hypothesis tests of the mean. For a two-tailed hypothesis test with level of significance $\alpha$ and null hypothesis $H_{0}$ : $\mu=k,$ we reject $H_{0}$ whenever $k$ falls outside the $c=1-\alpha$ confidence interval for $\mu$ based on the sample data. When $k$ falls within the $c=1-\alpha$ confidence interval, we do not reject $H_{0}$ (A corresponding relationship between confidence intervals and two-tailed hypothesis tests also is valid for other parameters, such as $p, \mu_{1}-\mu_{2},$ and $p_{1}-p_{2},$ which we will study in Sections 8.3 and $8.5 .$ ) Whenever the value of $k$ given in the null hypothesis falls outside the $c=1-\alpha$ confidence interval for the parameter, we reject $H_{0} .$ For example, consider a two-tailed hypothesis test with $\alpha=0.01$ and $$H_{0}: \mu=20 \quad H_{1}: \mu \neq 20$$ A random sample of size 36 has a sample mean $\bar{x}=22$ from a population with standard deviation $\sigma=4.$ (a) What is the value of $c=1-\alpha ?$ Using the methods of Chapter $7,$ construct a $1-\alpha$ confidence interval for $\mu$ from the sample data. What is the value of $\mu$ given in the null hypothesis (i.e., what is $k$ )? Is this value in the confidence interval? Do we reject or fail to reject $H_{0}$ based on this information? (b) Using methods of this chapter, find the $P$ -value for the hypothesis test. Do we reject or fail to reject $H_{0}$ ? Compare your result to that of part (a).
Hypothesis Testing
Testing the Mean $\mu$
Recommended Textbooks
Elementary Statistics a Step by Step Approach
The Practice of Statistics for AP
Introductory Statistics
18,000,000+
Students on Numerade
Trusted by students at 8,000+ universities
Watch the video solution with this free unlock.
EMAIL
PASSWORD