Bruce, a store owner, would like to determine if a new advertising initiative has increased his sales per day compared to last year. He gathers information on 14 random sales days and conducts a hypothesis test about the sales revenue during a day using a significance level of 0.5 %. The mean revenue per day prior to the initiative was 650 dollars. df t0.10 t0.05 t0.025 t0.01 t0.005 ... ... ... ... ... ...13 1.350 1.771 2.160 2.650 3.01214 1.345 1.761 2.145 2.624 2.99715 1.341 1.753 2.131 2.602 2.947 16 1.337 1.746 2.120 2.583 2.921 Determine the critical value(s) using the partial t-table above. If entering two critical values, use +-.

The critical value of t =3.012

Step-by-step explanation:

Solution

Recall that:

Bruce a store owner put together information on random sales of = 14 days

A significant level of =0.5%

The mean revenue per day = 65 dollars

Now,

We find he critical values using the partial t-table stated above

Thus,

The degree of freedom is given below:

n -1 = 14

14-1 =13

For a 0.005 level and right tailed test and with a 13 degree of freedom, the critical value of t =3.012.

an advantage of the standard deviation is that it increases as the dispersion of the data increases. e. the interquartile range is preferred when the data are not skewed or no have outliers. an advantage of the standard deviation is that it uses all the observations in its computation.