Question

Use technology to find the​ P-value for the hypothesis test described below. The claim is that for 12 AM body​ temperatures, the mean is μ > 98.6°F. The sample size is n = 6 and the test statistic is t = 1.965.

Answer 1

Answer:

$p_v =P(t_5>1.965)=0.0533$

We can use the following excel code to find it :"=1-T.DIST(1.965,5,TRUE) "

Step-by-step explanation:

Data given and notation  

$\bar X$ represent the average score for the sample  

$s$ represent the sample standard deviation  

$n=6$ sample size  

$\mu_o =98.6$ represent the value that we want to test  

$\alpha$ represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

$p_v$ represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to apply a one upper tailed test.  

What are H0 and Ha for this study?  

Null hypothesis: $\mu \leq 98.6$  

Alternative hypothesis :$\mu > 98.6$  

Compute the test statistic  

The statistic for this case is given by:  

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)  

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic  

The calculated value on this case is given t=1.965

Give the appropriate conclusion for the test  

First we need to calculat ethe degrees of freedom given by:

$df=n-1=6-1=5$

Since is a one side right tailed test the p value would be:  

$p_v =P(t_5>1.965)=0.0533$

We can use the following excel code to find it :"=1-T.DIST(1.965,5,TRUE) "

Conclusion  

If we compare the p value and a significance level assumed $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the true mean is not significantly higher than 98.6 at 5% of significance.