Answer:
a) Null hypothesis:
Alternative hypothesis:
b)
The degrees of freedom are given by:
![df =n-1= 51-1=50](https://tex.z-dn.net/?f=%20df%20%3Dn-1%3D%2051-1%3D50)
Now we can calculate the p value taking in count the alternative hypothesis
Since the p value is lower than the significance
we have enough evidence to reject the null hypothesis and the true mean is significantly different from 14.44
c) ![12.50 -2.01 \frac{4.75}{\sqrt{51}}= 11.163](https://tex.z-dn.net/?f=%2012.50%20-2.01%20%5Cfrac%7B4.75%7D%7B%5Csqrt%7B51%7D%7D%3D%2011.163)
![12.50 +2.01 \frac{4.75}{\sqrt{51}}= 13.837](https://tex.z-dn.net/?f=%2012.50%20%2B2.01%20%5Cfrac%7B4.75%7D%7B%5Csqrt%7B51%7D%7D%3D%2013.837)
Step-by-step explanation:
Information given
represent the mean for the daily iron intake
represent the sample deviation
sample size
represent the reference value
represent the significance level for the hypothesis test.
t would represent the statistic
represent the p value for the test
Part a
We want to test if the mean iron intake among the low-income group is different from that of the general population, the system of hypothesis would be:
Null hypothesis:
Alternative hypothesis:
Part b
Since we don't know the population deviation the statistic would be given by:
(1)
Replacing the info we got
The degrees of freedom are given by:
![df =n-1= 51-1=50](https://tex.z-dn.net/?f=%20df%20%3Dn-1%3D%2051-1%3D50)
Now we can calculate the p value taking in count the alternative hypothesis
Since the p value is lower than the significance
we have enough evidence to reject the null hypothesis and the true mean is significantly different from 14.44
Part c
The confidence interval would be given by:
![\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}](https://tex.z-dn.net/?f=%5Cbar%20X%20%5Cpm%20t_%7B%5Calpha%2F2%7D%5Cfrac%7Bs%7D%7B%5Csqrt%7Bn%7D%7D)
For the 95% confidence interval we can find the critical value in a t distribution with 50 degrees of freedom and we got:
![t_{\alpha/2}= 2.01](https://tex.z-dn.net/?f=t_%7B%5Calpha%2F2%7D%3D%202.01)
And replacing we got:
![12.50 -2.01 \frac{4.75}{\sqrt{51}}= 11.163](https://tex.z-dn.net/?f=%2012.50%20-2.01%20%5Cfrac%7B4.75%7D%7B%5Csqrt%7B51%7D%7D%3D%2011.163)
![12.50 +2.01 \frac{4.75}{\sqrt{51}}= 13.837](https://tex.z-dn.net/?f=%2012.50%20%2B2.01%20%5Cfrac%7B4.75%7D%7B%5Csqrt%7B51%7D%7D%3D%2013.837)