 Amber Quinn

2022-06-21

What is the p value if the test static is 1.319 and the significance level is 0.01? pheniankang

Testing of Hypothesis:
Testing of hypothesis is a rule which, when the sample values have been obtained, leads to a decision to accept or to reject the hypothesis under consideration.
Level of Significance:
In testing of hypothesis, Level of Significance is the probability of rejecting a null hypothesis when it is true. In other words, if the value of test statistic exceeds the critical value, we conclude that the null hypothesis has been rejected.
p-Value:
In hypothesis testing, the p-value is defined as the probability of getting results at least as extreme as the observed results. The p-value is used to provide the smallest level of significance at which the null hypothesis would be rejected. Smaller the p-value, stronger the evidence in favor of the alternative hypothesis.
Here, we are given the value of the test statistic but nothing has been mentioned about the degrees of freedom. So, the test is not a t-test, chi-square test or F-test.
So, the test statistic is of a z-test.
For a z-test, the p-value can be obtained as follows:
(for an one-sided test)
p-value is calculated as the probability that the z-value is greater than the value of the test statistic (for right-tailed test) or the probability that the z-value is less than the value of the test statistic (for left-tailed test).
$p-value=P\left(Z>1.319\right)\phantom{\rule{0ex}{0ex}}=-P\left(Z\le 1.319\right)\phantom{\rule{0ex}{0ex}}=-0.9064\text{[from stadart normal distribution table]}\phantom{\rule{0ex}{0ex}}=0..0936\phantom{\rule{0ex}{0ex}}\text{The significance level}\phantom{\rule{0ex}{0ex}}\alpha =0.01\phantom{\rule{0ex}{0ex}}\text{Clearly,}\phantom{\rule{0ex}{0ex}}p>\alpha$
Hence, the test is not significance.
(for a two-sided test)
p-value is calculated as the probability that the z-value is greater than the value of the test statistic or less than the negative value of the test statistic.
$p-value=P\left(Z>1.319\right)+P\left(Z<-1.319\right)\phantom{\rule{0ex}{0ex}}=1-P\left(Z\le 1.319\right)+P\left(Z<-1.319\right)\phantom{\rule{0ex}{0ex}}=1-0.9064+0.0936\text{[from stadart normal distribution table]}\phantom{\rule{0ex}{0ex}}=0.1872\phantom{\rule{0ex}{0ex}}\text{The significance level}\phantom{\rule{0ex}{0ex}}\alpha =0.01\phantom{\rule{0ex}{0ex}}\text{Clearly,}\phantom{\rule{0ex}{0ex}}p>\alpha$
Hence, the test is not significance.

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