A research center claims that more than \( 25 \% \) of employees in a certain country have changed jobs in the past three years. In a random sample of 260 people from that country, 78 have changed jobs in the past three years: At \( \alpha=0.05 \), is there enough evidence to support the center's claim? Complete parts (a) through (d) below. (b) Find the critical value(s) and identify the rejection region(s). Identify the critical value(s) for this test. \( z_{0}=\square \) (Round to two decimal places as needed. Use a comma to separate answers as needed.) Identify the rejection region(s). Select the correct choice below and fill in the answer box(es) to complete your choice. The rejection region is \( \square<z<\square \). (Round to two decimal places as needed.) B. The rejection regions are \( z<\square \) and \( z>\square \). C. The rejection region is \( z>\square \). D. The rejection region is \( z<\square \). (c) Find the standardized test statistic \( z \).

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).

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)

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? \( x=\square \)

21. Una empresa de investigación de mercado emplea una muestra de individuos para calificar el potencial de compra de un determinado producto antes y después de que los individuos vean un comercial de televisión acerca del mismo. La calificación del potencial de compra se hace con una escala del 0 al 10 , con los valores más altos indicando un mayor potencial de compra. En la hipótesis nula se establece que la media de las calificaciones de "después" será menor o igual a la media de las calificaciones "antes". El rechazo de esta hipótesis indica que el comercial me- jöra la media de la calificación al potencial de compra. Use \( \alpha=0.05 \) y los datos de la tabla si- guiente para probar esta hipótesis y haga un comentario sobre la utilidad del comercial. Calificación al potencial de compra Individuos Después 1

Baseball regression line prediction: Suppose the regression line for the number of runs scored in a season, \( y \), is given by \( \hat{y}=-730+4300 x \), where \( x \) is the team's batting average. a. For a tearn with a batting average of 0.235 , find the expected number of runs scored in a season. Round your answer to the nearest whole number. b. If we can expect the number of runs scored in a scason is 480 , then what is the assumed team's batting average? Round your onswer to three deeima! places.