Answer :
Answer:
a) Null hypothesis:[tex]\mu = 140[/tex]
Alternative hypothesis:[tex]\mu \neq 140[/tex]
b) [tex]t=\frac{170-140}{\frac{80}{\sqrt{64}}}=3[/tex]
c) The degrees of freedom are given by:
[tex]df=n-1=64-1=63[/tex]
Now we can calculate the p value, since we are conducting a two tailed test:
[tex]p_v =2*P(t_{63}>3)=0.0039[/tex]
Since the p value is lower than the significance level of [tex]\alpha=1-0.95=0.05[/tex] we have enough evidence to conclude that the true mean is significantly different from the US average of 140 minutes
Step-by-step explanation:
Information provided
[tex]\bar X=170[/tex] represent the sample mean in minutes
[tex]s=80[/tex] represent the standard deviation
[tex]n=64[/tex] sample size
[tex]\mu_o =140[/tex] represent the value to verify
[tex]\alpha[/tex] represent the significance level
t would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value
Part a
We are interested in determining whether or not the average usage in Soddy-Daisy is significantly different from the U.S. average (140 minutes), the system of hypothesis would be:
Null hypothesis:[tex]\mu = 140[/tex]
Alternative hypothesis:[tex]\mu \neq 140[/tex]
Part b
Since we don't know the population deviation 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{170-140}{\frac{80}{\sqrt{64}}}=3[/tex]
Part c
The degrees of freedom are given by:
[tex]df=n-1=64-1=63[/tex]
Now we can calculate the p value, since we are conducting a two tailed test:
[tex]p_v =2*P(t_{63}>3)=0.0039[/tex]
Since the p value is lower than the significance level of [tex]\alpha=1-0.95=0.05[/tex] we have enough evidence to conclude that the true mean is significantly different from the US average of 140 minutes