In a lab experiment, the decay of a radioactive isotope is being observed. At the beginning of the first day of the experiment t
2 answers:
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Answer:
(a) The value = 72.3 is 1.21 standard deviations below the null value of <em>μ</em>.
(b) The mean drying time of paint is not less than 74.
Step-by-step explanation:
The hypothesis for the single mean test is:
<em>H₀</em>: <em>μ</em> = 74 vs. <em>Hₐ</em>: μ < 74.
The information provided is:
<em>σ</em> = 7
<em>n</em> = 25
As the population standard deviation is known, we will use a <em>z</em>-test for single mean.
(a)
The <em>z</em>-score is a Normal distribution with mean 0 and variance 1. It is also defined as the number of standard deviations a raw score is from the mean.
The <em>z</em>-score for sample mean is given by:
If = 72.3 compute the corresponding <em>z</em>-score as follows:
Thus, the value = 72.3 is 1.21 standard deviations below the null value of <em>μ</em>.
(b)
So the test statistic is, <em>z</em> = -1.21.
Compute the <em>p</em>-value of the test as follows:
*Use a <em>z</em>-table.
The decision rule says to, reject the null hypothesis if the <em>p</em>-value is less than the significance level <em>α</em> = 0.01 and vice-versa.
<em>p</em>-value = 0.1131 < <em>α</em> = 0.01
The null hypothesis was failed to be rejected.
Thus, the mean drying time of paint is not less than 74.