Use a \( \chi^{2} \)-test to test the claim \( \sigma<44 \) at the \( \alpha=0.05 \) significance level using sample statistics \( s=40.3 \) and \( n=1 \). Assume the population is normally distributed. \( H_{0}: \sigma \leq 44 \) Identify the standardized test statistic. 15.10001 (Round to three decimal places as needed.) Identify the critical value(s). \( \square \) (Round to three decimal places as needed. Use a comma to separate answers as needed.) Choose the correct conclusion below.

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

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)

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

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? \( x=\square \)