Answer:
t=5.5080( to 3 d.p)
Step-by-step explanation:
From the data given,
n =20
Deviation= 34/20= 1.7
Standard deviation (sd)= 1.3803(√Deviation)
Standard Error = sd/√n
= 1.3803/V20 = 0.3086
Test statistic is:
t = deviation /SE
= 1.7/0.3086 = 5.5080
ndf = 20 - 1 = 19
alpha = 0.01
One Tailed - Right Side Test
From Table, critical value of t =2.5395
Since the calculated value of t = 5.5080 is greater than critical value of t = 2.5395, the difference is significant. Reject null hypothesis.
t score = 5.5080
ndf = 19
One Tail - Right side Test
By Technology, p - value = 0.000
Since p - value is less than alpha , reject null hypothesis.
Conclusion:
From the result obtained it can be concluded that ,the data support the claim that the mean rating assigned to the wine when the cost is described as $90 is greater than the mean rating assigned to the wine when the cost is described as $10.
1). y-3=-2(x+4)
2) slope = 3/4
y+2=3/4(x+4)
y=3/4x +1
3x-4y=4
looks like only the last one
170,000 square kilometers, divide by 10, or take off a 0
Answer:
Yes, *2
Step-by-step explanation:
The second figure is exactly half of the first figure
