Respuesta :
Answer:
The P-value is 0.3015 and the conclusion is to reject the null hypothesis.
Step-by-step explanation:
We are given the level of significance of 0.05 and the test statistic in a right-tailed test is z = 0.52.
Now, the decision rule to reject or fail to reject the null hypothesis based on the P-value is given as;
- If the P-value of the test statistics is less than the level of significance, then we have sufficient evidence to reject our null hypothesis due to which we conclude to reject the null hypothesis.
- If the P-value of the test statistics is more than the level of significance, then we have insufficient evidence to reject our null hypothesis due to which we fail to reject the null hypothesis.
SO, P-value of the test statistics is calculated as;
P-value = P(Z > 0.52) = 1 - P(Z [tex]\leq[/tex] 0.52)
= 1 - 0.6985 = 0.3015
Now, as the P-value of the test statistics is more than the level of significance as 0.3015 > 0.05, so we have sufficient evidence to reject our null hypothesis due to which we conclude to reject the null hypothesis.
Using the z-distribution, we have that:
- The p-value of the test is of 0.3015.
- Since the p-value is greater than the significance level, we fail to reject the null hypothesis.
We have a right-tailed test, hence we are testing if the mean is greater than a value, then the p-value is 1 subtracted by the p-value of z = 0.52.
- Looking at the z-table, z = 0.52 has a p-value of 0.6985.
1 - 0.6985 = 0.3015
The p-value of the test is of 0.3015.
Since the p-value is greater than the significance level, we fail to reject the null hypothesis.
A similar problem is given at https://brainly.com/question/24166849
