Respuesta :
Answer:
[tex] t= \frac{1.743-1.73}{\frac{0.107}{\sqrt{30}}}= 0.665[/tex]
[tex] df = n-1= 30-1=29[/tex]
[tex] p_v= P(t_{29}>1.73)= 0.047[/tex]
And if we compare the p value obtained and the significance level provided of 0.05 or 5% we see that [tex]p_v<\alpha[/tex] so then we have enough evidence to conclude that the true mean for this case is significantly higher tha 1.73 since we reject the null hypothesis for the test
Step-by-step explanation:
We have the following dataset given:
1.7, 1.6, 1.8, 1.9, 1.75, 1.83, 1.82, 1.65, 1.95, 1.69, 1.82, 1.87, 1.65, 1.54, 1.98, 1.78, 1.69, 1.75, 1.62, 1.64, 1.75, 1.8, 1.65, 1.7, 1.82, 1.62, 1.83, 1.75, 1.7, 1.65
We can calculate the sample mean and deviation from this data with the following 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]
And replacing we got:
[tex]\bar X= 1.743 ,s=0.107 , n=30[/tex]
From previous info we know that 10 years before, the average height of persons in country A was determined to be 1.73 and standard deviation is 0.2.
We can use a one t test in order to check the claim that he average height of individuals in country A has increased in the last 10 years
Null hypothesis: [tex]\mu \leq 1.73[/tex]
Alternative hypothesis: [tex]\mu>1.73[/tex]
The statistic is given by:
[tex] t=\frac{\bar X \mu}{\frac{s}{\sqrt{n}}}[/tex]
Replacing the info given we got:
[tex] t= \frac{1.743-1.73}{\frac{0.107}{\sqrt{30}}}= 0.665[/tex]
The degrees of freedom for this test are given by:
[tex] df = n-1= 30-1=29[/tex]
Now we can calculate the p value with the following probability:
[tex] p_v= P(t_{29}>1.73)= 0.047[/tex]
And if we compare the p value obtained and the significance level provided of 0.05 or 5% we see that [tex]p_v<\alpha[/tex] so then we have enough evidence to conclude that the true mean for this case is significantly higher tha 1.73 since we reject the null hypothesis for the test
