Answer:
[tex]t=\frac{46-40}{\frac{5.148}{\sqrt{5}}}=2.606[/tex]
The degrees of freedom are given by:
[tex]df=n-1=5-1=4[/tex]
The p value wuld be given by:
[tex]p_v =2*P(t_{(4)}>2.606)=0.060[/tex]
For this case the p value is higher than the significance level so then we can conclude that the true mean is not significantly different from 40
The distribution with the critical values are in the figure attached
Step-by-step explanation:
Information given
48, 41, 40, 51, and 50
The sample mean and deviation can be calculated with these formulas:
[tex]\bar X= \frac{\sum_{i=1}^n X_i}{n}[/tex]
[tex]s =\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}[/tex]
[tex]\bar X=46[/tex] represent the mean height for the sample
[tex]s=5.148[/tex] represent the sample standard deviation
[tex]n=5[/tex] sample size
[tex]\mu_o =40[/tex] represent the value that we want to test
[tex]\alpha=0.05[/tex] represent the significance level for the hypothesis test.
t would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value for the test
Hypothesis to test
We want to test if the true mean for this case is equal to 40, the system of hypothesis would be:
Null hypothesis:[tex]\mu = 40[/tex]
Alternative hypothesis:[tex]\mu \neq 40[/tex]
The statistic is given by:
[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)
Replacing we got:
[tex]t=\frac{46-40}{\frac{5.148}{\sqrt{5}}}=2.606[/tex]
The degrees of freedom are given by:
[tex]df=n-1=5-1=4[/tex]
The p value wuld be given by:
[tex]p_v =2*P(t_{(4)}>2.606)=0.060[/tex]
For this case the p value is higher than the significance level so then we can conclude that the true mean is not significantly different from 40
The distribution with the critical values are in the figure attached
