Respuesta :
Answer:
The one tailed P-value for the test statistic for df = n-1 = 33-1 = 32 is
[tex]P-value=0.1575[/tex]
Step-by-step explanation:
We are given:
[tex]t=1.021[/tex]
[tex]n=33[/tex]
Claim: [tex]\mu>98.6[/tex]°F
Now we can use the technology like excel to find the P-value. The excel function is given below:
[tex]=TDIST(1.021,32,1)=0.1575[/tex]
Where:
1.021 is the test statistic
32 is the degrees of freedom
1 stands for one tailed alternative hypothesis.
Therefore, the P-value is 0.1575
Answer:
The required [tex]P-[/tex]value using technology is [tex]0.1575[/tex]
Step-by-step explanation:
Given: The claim is that for [tex]12[/tex] AM body temperatures, the mean is [tex]\mu>98.698, \; 6^\circ\;F[/tex]. The sample size is [tex]n=33[/tex] and the test statistic is [tex]t=1.021[/tex].
The one tailed [tex]P-[/tex]value for the test statistic for [tex]df=n-1=33-1=32[/tex] is calculated as:
Now, using the technology like excel to find the [tex]P-[/tex]value of function. The excel function is given below:
[tex]TDIST(1.021,32,1)=0.1575[/tex]
[tex]\rm{Where}\\1.021\; \rm{is \; the\; \; test \;statistic}\\32\; \rm{is\; the \;degrees \;of \;freedom}\\1 \;\rm{stands \;for\; one\; tailed \;alternative\; hypothesis}[/tex]
Therefore, the required [tex]P-[/tex]value is [tex]0.1575[/tex].
Learn more about P-value here:
https://brainly.com/question/24343388?referrer=searchResults
