The mean weight of a pair of shoes is 878.00 g with a standard deviation of 49.00 g . The mean weight of an empty shoe box for a pair of shoes is 73.00 g with a standard deviation of 6.60 g . What are the mean and standard deviation of the weight of a shoe box with a pair of shoes inside? Mean = [?] S.D. \( =[\square \)

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? \( x=\square \)

What is a scatter plot?

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).