A scientist conducts an experiment to determine if the mean…

A scientist conducts an experiment to determine if the mean alkalinity level of water specimens from the Olentangy River is greater than 50 milligrams per liter (mpl).  She selects a random sample of 100 water specimens from the river and finds a sample mean of 67.8 mpl and a sample standard deviation of 14.4 mpl.  She decides to test the hypothesis using a significance level of 0.01.  Using this information, calculate the value of the test statistic.

A two-factor factorial experiment is conducted to compare li…

A two-factor factorial experiment is conducted to compare litter sizes of Yorkshire and Landrace sows derived either from a line unselected for litter size or from a line that has gone through 15 years of selection for increased litter size.  Two sows of each breed are randomly selected from each line.  Their litter sizes are as follows:                                   Yorkshire       Landrace            Unselected line               8                     9                                        9                    10 Selected line                 11                    11                                      10                      9                                                                                     The partially completed ANOVA table is as follows:  Source             df             SS                MS            F  Total                              7.875 Line                               3.125            3.125       3.57 Breed Line x Breed Error                              3.500            0.875                                                                                       Find the mean squares for breed.

Assume that the probabilities of two genetic defects (we wil…

Assume that the probabilities of two genetic defects (we will call them defect A and defect B) in horses are 0.05 and 0.08, respectively.  If these two genetic defects represent independent events, what is the probability that a horse will have both of these genetic defects?

A statistician evaluated the winning strategies of teams in…

A statistician evaluated the winning strategies of teams in the National Football League (NFL).  He used actual NFL play-by-play data to approximate the probabilities associated with certain outcomes (e.g., running plays, short pass plays, and long pass plays).  The table below shows the probability distribution for the yardage gained, X, on a running play.  A negative gain represents a loss of yards on the play. Find the probability of gaining 10 or more yards on a running play. X, Yards Gained            Probability                -4                                  0.020        -2                                  0.060        -1                                  0.070          0                                 0.150          1                                 0.130          2                                 0.110          3                                 0.090          4                                 0.070          6                                 0.090          8                                 0.060         10                                0.050         15                                0.085         30                                0.010         50                                0.004         99                                0.001                                                               

A Gallop poll is conducted to estimate the proportion of vot…

A Gallop poll is conducted to estimate the proportion of voters who plan to vote in favor of a school levy in a certain school distruct.  A random sample of 400 people of voting age is selected.  Results of the poll show that 240 of the 400 people polled plan to vote in favor of the school levy.  What is the point estimate of the true population proportion of people who plan to vote in favor of the school levy?

A two-factor factorial experiment is conducted to compare li…

A two-factor factorial experiment is conducted to compare litter sizes of Yorkshire and Landrace sows derived either from a line unselected for litter size or from a line that has gone through 15 years of selection for increased litter size.  Two sows of each breed are randomly selected from each line.  Their litter sizes are as follows:                                   Yorkshire       Landrace            Unselected line               8                     9                                        9                    10 Selected line                 11                    11                                      10                      9                                                                                     The partially completed ANOVA table is as follows:  Source             df             SS                MS            F  Total                              7.875 Line                               3.125            3.125       3.57 Breed Line x Breed Error                              3.500            0.875                                                                                       What is the correct conclusion regarding the line x breed interaction?  Use a significance level of 0.05.

Suppose that a 95% confidence interval for μ turns out to be…

Suppose that a 95% confidence interval for μ turns out to be (100 lb, 500 lb).  To make more useful inferences from the data, it is desired to reduce the width of the confidence interval.  What action should we take to reduce the width of the confidence interval?

A two-factor factorial experiment is conducted to compare fl…

A two-factor factorial experiment is conducted to compare fleece weights of Merino, Suffolk, and Dorset ewes fed one of two diets.  Two ewes of each breed are randomly assigned to each diet.  The fleece weights (in pounds) are as follows:                Merino            Suffolk            Dorset        Diet 1          14                   9                    8                     15                 10                    8 Diet 2           13                   8                  11                     12                    9                 12                                                                                   The partially completed ANOVA table is as follows: Source             df           SS             MS           F      Total                          66.2500 Diet                             0.0833       0.0833       0.19992 Breed                        46.5000     23.2500     55.79955 Diet x Breed              17.1667     Error                            2.5000      0.41667                                                                                       The correct values for the diet x breed mean squares and F value, respectively, are: