A sample of 39 observations is selected from a normal population. The sample mean is 19, and the population standard deviation is 4. Conduct the following test of hypothesis using the 0.10 significance level.
H_{0}: μ ≤ 18
H_{1}: μ > 18
One-tailed test
Two-tailed test
Reject H_{0} when z > 1.282
Reject H_{0} when z ≤ 1.282
Reject H_{0}
Fail to reject H_{0}
a)
One-tailed test
b)
Reject H_{0} when z > 1.282
c)
population mean μ= | 18 | ||
sample mean 'x̄= | 19.000 | ||
sample size n= | 39 | ||
std deviation σ= | 4.00 | ||
std error ='σx=σ/√n=4/√39= | 0.641 | ||
z test statistic= ='(x̄-μ)/σx=(19-18)/0.641= | 1.56 |
d)
since test statistic falls in rejection region we reject null hypothesis |
Reject H_{0}
e-1)
p value = | 0.0594 | (from excel:1*normsdist(-1.56) |
e-2_)
p value is probability of getting as or more extreme test statistic if null hypothesis is true(population mean is 18).
.
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