Consider the following hypotheses:
H_{0}: μ = 36
H_{A}: μ ≠ 36
The population is normally distributed. A sample produces the
following observations: (You may find it useful to
reference the appropriate table: z table
or t table)
38 | 36 | 38 | 41 | 38 | 38 | 39 |
Click here for the Excel Data File
a. Find the mean and the standard deviation. (Round your answers to 2 decimal places.)
b. Calculate the value of the test statistic.
(Round intermediate calculations to at least 4 decimal
places and final answer to 2 decimal places.)
c. Find the p-value.
p-value < 0.01
d. At the 5% significance level, what is the
conclusion?
Do not reject H_{0} since the p-value is smaller than α.
Do not reject H_{0} since the p-value is greater than α.
Reject H_{0} since the p-value is smaller than α.
Reject H_{0} since the p-value is greater than α.
e. Interpret the results αα =
0.05.
We cannot conclude that the sample mean differs from 36.
We conclude that the population mean differs from 36.
We cannot conclude that the population mean differs from 36.
We conclude that the sample mean differs from 36.
Solution:
Here, we have to use one sample t test for the population mean.
The null and alternative hypotheses are given as below:
H0: µ = 36 versus Ha: µ ≠ 36
This is a two tailed test.
Part a
From given data, we have
µ = 36
Xbar = 38.29
S = 1.50
n = 7
df = n – 1 = 6
α = 0.05
Critical value = - 2.4469 and 2.4469
(by using t-table or excel)
Part b
The test statistic formula is given as below:
t = (Xbar - µ)/[S/sqrt(n)]
t = (38.28571429 – 36)/[ 1.496026483/sqrt(7)]
t = 4.0423
Part c
P-value = 0.0068
(by using t-table)
p-value < 0.01
Part d
P-value < α = 0.05
So, we reject the null hypothesis
Reject H0 since the p-value is smaller than α.
Part e
We conclude that the population mean differs from 36.
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