Respuesta :
Answer:
[tex]z=\frac{10.8-10}{\frac{2.4}{\sqrt{36}}}=2[/tex]
Now we can calculate the p value since we are using ta right tailed test
[tex]p_v =P(z>2)=0.0228[/tex]
Since the p value is lower than the significance level provided of 0.05 we have enough evidence to reject the null hypothesis and we can conclude that the members of the Priority Health Club lose an average of 10 pounds or more within the first month after joining the club
Step-by-step explanation:
Data provided
[tex]\bar X=10.8[/tex] represent the sample eman for the amount of weigth lost
[tex]\sigma = 2.4[/tex] represent the population standard deviation
[tex]n=36[/tex] sample size
[tex]\mu_o =10[/tex] represent the value to test
[tex]\alpha=0.05[/tex] represent the significance level for the hypothesis test.
z would represent the statistic
[tex]p_v[/tex] represent the p value for the test
Hypothesis to test
We want to determine if the members of the Priority Health Club lose an average of 10 pounds or more within the first month after joining the club , the system of hypothesis would be:
Null hypothesis:[tex]\mu \leq 10[/tex]
Alternative hypothesis:[tex]\mu > 10[/tex]
Since we know the population deviation the statistic can be founded like this:
[tex]z=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)
Replacing theinfo given we got:
[tex]z=\frac{10.8-10}{\frac{2.4}{\sqrt{36}}}=2[/tex]
Now we can calculate the p value since we are using ta right tailed test
[tex]p_v =P(z>2)=0.0228[/tex]
Since the p value is lower than the significance level provided of 0.05 we have enough evidence to reject the null hypothesis and we can conclude that the members of the Priority Health Club lose an average of 10 pounds or more within the first month after joining the club
