Given the following hypothesis: |
H_{0} : μ = 125 |
H_{1} : μ ≠ 125 |
A random sample of six resulted in the following values 132, 132, 130, 143, 140, and 130. |
Using the 0.05 significance level, can we conclude the mean is different from 125? |
(a) | What is the decision rule? (Negative amounts should be indicated by a minus sign. Round your answers to 3 decimal places.) |
Reject H_{0} : μ = 125 and fail to reject H_{1} : μ ≠ 125 when the test statistic is (Click to select)inside the intervaloutside the interval ( , ). |
(b) | The value of the test statistic is (Round your answer to 3 decimal places.) |
(c) | What is your decision regarding H_{0} ? |
(Click to select)Fail to rejectReject |
(d) | Estimate the p-value. |
(Click to select)Between 0.01 and 0.05Less than 0.01Greater than 0.1Between 0.05 and 0.2 |
Part a
Here, we have to use one sample t test for the population mean.
The null and alternative hypotheses are given as below:
H0: µ = 125 versus Ha: µ ≠ 125
This is a two tailed test.
The test statistic formula is given as below:
t = (Xbar - µ)/[S/sqrt(n)]
From given data, we have
µ = 125
Xbar = 134.5
S = 5.576737397
n = 6
df = n – 1 = 5
α = 0.05
Critical value = - 2.5706 and 2.5706
(by using t-table or excel)
Reject H0 : μ = 125 and fail to reject H1 : μ ≠ 125 when the test statistic is outside the interval (-2.571, 2.571). |
Part b
t = (Xbar - µ)/[S/sqrt(n)]
t = (134.5 – 125)/[ 5.576737397/sqrt(6)]
t = 4.1727
Value of the test statistic = 4.173
Part c
Value of the test statistic is outside the above critical values, so we reject H0.
Part d
P-value = 0.0087
(by using t-table)
(Less than 0.01)
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