Spencer randomly samples the out-of-state tuition and fees for 25 public universities and found a mean of $21,032 with a standard deviation of $2,494. Using the alternative hypothesis that µ < $21,706, Spencer found a z test statistic of -1.35. Given the data provided, what is the p-value of the z-test statistic?

0.0885

Step-by-step explanation:

To find :  p-value of the z-test statistic.

p-value of the z-statistic depends on the alternative hypothesis.

Here , Alternative hypothesis :  µ < $21,706 , which is a left tailed hypothesis.

So we will check the area to the left side of the z-statistic.

Since z= -1.35

P(Z<z) =P(Z<-1.35)= 1- P(Z<1.35)

= 1- 0.9115

= 0.0885

Hence, the required p-value = 0.0885

step-by-step explanation:

circumference divided by 3.141592654 then timesthe radius twice

$2416.6

step-by-step explanation:

per hour = $ 25

total work time = 35

commission = 15%

total she has to make = $ 4,500

let she sold x dollar products

so her commission will be 15 % of x

now

total = weekly basic pay + commission

4500= (25*35) + 15 % of x

4500=875 + * x

4500-875 =   * x

3625= * x

multiplying wwith 100 on both sides gives

36250 = 15 x

now

x =  

x= $2416.6