If LaTeX: z_{obt}z o b t= +2.21, do you reject LaTeX: H_0H 0 or fail to reject LaTeX: H_0H 0 using a one-tailed test in the uppe
1 answer:
Answer:
Reject the null hypothesis
Step-by-step explanation:
From the question we are told that
The test statistic is
The level of significance is
Given that this a one tailed in the upper tail then from the z table the area under the normal curve to the right corresponding to is
So the p-value is
From the values obtained we see that hence
The decision rule
Reject the null hypothesis
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I got D.
There's a few ways to solve it; I prefer using tables, but there are functions on a TI-84 that'll do it for you too. The logic here is, you have a standard normal distribution which means right away, the mean is 0 and the standard deviation is 1. This means you can use a Z table that helps you calculate the area beneath a normal curve for a range of values. Here, your two Z scores are -1.21 and .84. You might notice that this table doesn't account for negative values, but the cool thing about a normal distribution is that we can assume symmetry, so you can just look for 1.21 and call it good. The actual calculation here is:
1 - Z-score of 1.21 - Z-score .84 ... use the table or calculator
1 - .1131 - .2005 = .6864
Because this table calculates areas to the RIGHT of the mean, you have to play around with it a little to get the bit in the middle that your graph asks for. You subtract from 1 to make sure you're getting the area in the middle and not the area of the tails in this problem.
