For the following information, determine whether a normal sampling distribution can be used, where \( p \) is the population proportion, \( \alpha \) is the level of significance, \( \hat{p} \) is the sample proportion, and \( n \) is the sample size. If it can be used, test the claim. Claim: \( p \geq 0.47 ; \alpha=0.04 \). Sample statistics: \( \hat{p}=0.40, n=140 \) If a normal sampling distribution can be used, identify standardized test statistic \( z \). Select the correct choice below and, if necessary, fill in the answer box to complete your choice. (Round to two decimal places as needed.) B. A normal sampling distribution cannot be used. If a normal sampling distribution can be used, decide whether to reject or fail to reject the null hypothesis and interpret the decision. Choose the correct answer below. A. Reject the null hypothesis. There is not enough evidence to reject the claim. B. Fail to reject the null hypothesis. There is not enough evidence to reject the claim. C. Fail to reject the null hypothesis. There is enough evidence to reject the claim. D. Reject the null hypothesis. There is enough evidence to reject the claim. E. A normal samoling distrihution cannot he used

Why is data collection appropriate for this two-sample proportion procedure? Because the order of applying the two treatments to each doctor was randomized in this matched pairs design. Because two separate simple random samples were taken-one from each population. Because the doctors were randomly allocated to the two treatments (aspirin and placebo).

You run a regression analysis on a bivariate set of data ( \( n=88 \) ), You obtain the regression equation \[ y=0.231 x-8.928 \] with a correlation confficient of \( r=0.38 \) (which is significant at \( \alpha=0.01 \) ), You want to predict what value (on average) for the explanatory variable will give you a value of 90 on the response variable. What is the predicted explanatory value? \( x=\square \)

With \( P \)-value \( =0.000 \) and \( \alpha=0.05 \), should we reject \( H_{0} \) ? yes no

A regression was run to determine if there is a relationship between hours of TV watched per day \( (\mathrm{x}) \) and number of situps a person can do \( (y) \). The results of the regression were: \( \begin{array}{l}\mathrm{y}=\mathrm{ax}+\mathrm{h} \\ \mathrm{a}=2.389 \\ \mathrm{~b}=37.535 \\ \mathrm{r}^{2}=0.412164 \\ \mathrm{r}=-0.642\end{array} \) Use this to predict the number of situps a person who watches 0.5 hours of TV can do (ta one decimal place)

Knowing that \( 99.7 \% \) of all \( z \)-scores are between -3 and +3 , what is the \( P \)-value for \( z=-5.14 \) ? 0.000 0.0514 0.2981