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.
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As determined by the graph, f(-9) = 9
Answer: Your answer would be 16y² + 12x - 4y
