Respuesta :
Answer:
[tex]t=\frac{5.2-5}{\frac{0.8}{\sqrt{40}}}=1.58[/tex]
The degrees of freedom are given by:
[tex]df=n-1=40-1=39[/tex]
Thep value for this case would be given by:
[tex]p_v =P(t_{(39)}>1.58)=0.061[/tex]
Since the p value is higher than the significance level of 0.05 we have enough evidence to FAIL to reject the null hypothesis and we can conclude that true mean is not significantly higher than 5.
Step-by-step explanation:
Information provided
[tex]\bar X=5.2[/tex] represent the sample mean
[tex]s=0.8[/tex] represent the sample standard deviation
[tex]n=40[/tex] sample size
[tex]\mu_o =5[/tex] represent the value to verify
[tex]\alpha=0.05[/tex] represent the significance level
t would represent the statistic
[tex]p_v[/tex] represent the p value
Hypothesis to test
We want to verify if the true mean is higher than 5, the system of hypothesis would be:
Null hypothesis:[tex]\mu \leq 5[/tex]
Alternative hypothesis:[tex]\mu > 5[/tex]
The statistic is given by:
[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)
Replacing the info given we got:
[tex]t=\frac{5.2-5}{\frac{0.8}{\sqrt{40}}}=1.58[/tex]
The degrees of freedom are given by:
[tex]df=n-1=40-1=39[/tex]
Thep value for this case would be given by:
[tex]p_v =P(t_{(39)}>1.58)=0.061[/tex]
Since the p value is higher than the significance level of 0.05 we have enough evidence to FAIL to reject the null hypothesis and we can conclude that true mean is not significantly higher than 5.