A sample of 36 observations is selected from a normal population. The sample mean is 21, and the sample stand?
H0: µ ≤ 20
H1: µ > 20
(a) Is this a one- or two-tailed test? (b) What is the decision rule? (c) What is the value of the test statistic? (d) What is your decision regarding H0? (e) What is the p-value? Interpret it.
(P.S. – The graph shows a z-critical of 1.65 instead of 1.645)
(a) Is this a one- or two-tailed test?
This is a one-tailed test, specifically an upper tail test.
(b) What is the decision rule?
Reject H0 if z > 1.645, otherwise accept H0.
(c) What is the value of the test statistic?
z-critical = 1.645
.5000 – .05 = .4500
Locate .4500 in the body of the z-table.
The z-value of 1.64 equals 0.4495 and
the z-value of 1.65 equals 0.4505,
so we split the difference and set z-crit to 1.645.
(d) What is your decision regarding H0?
z-calc = (xbar – µ) / (σ / sqrt(n))
xbar = sample mean (21)
µ = population mean (20)
σ = standard deviation of the population (5)
n = sample size (36)
z-calc = (21 – 20) / (5 / sqrt(36))
z-calc = (1) / (5 / 6)
z-calc = 1.2
Since z-calc of 1.2 < z-crit of 1.645, we accept H0. (e) What is the p-value? Interpret it. A z-score of 1.2 corresponds to 0.3849 The p-value is found by 0.5000 - 0.3849 = 0.1151. The probability of obtaining an x-bar greater than 21 if the population mean of 20 is 0.1151. The p-value of 0.1151 > .05 significance level.
Definition of the p-value:
The probability of observing a sample value as extreme as,
or more extreme than, the value observed, given that the null
hypothesis is true.
Good luck in your statistics class,
~ Mitch ~
P.S. – If you have a TI-83 or TI-84,
I **THINK** you can compute this directly on your calculator:
I believe it will also return a p-value also.
I’m able to do it on my TI-89, using a z-test.
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