Respuesta :
Answer:
a) The 95% confidence interval for the mean is (-1.171, 11.717).
b) No, there is not enough evidence to support the claim that increasing the amount of calcium in our diet reduce blood pressure (at a 5% significance level).
Step-by-step explanation:
The data for both groups are:
Group 1 (calcium): 7 -4 18 17 -3 -5 1 10 11 -2
Group 2 (placebo): -1 12 -1 -3 3 -5 5 2 -11 -1 -3
The mean and standard deviation for Group 1 is:
[tex]M_1=\dfrac{1}{10}\sum_{i=1}^{10}(7+(-4)+18+17+(-3)+(-5)+1+10+11+(-2))\\\\\\ M_1=\dfrac{50}{10}=5[/tex]
[tex]s_1=\sqrt{\dfrac{1}{(n-1)}\sum_{i=1}^{10}(x_i-M)^2}\\\\\\[/tex]
[tex]s_1=\sqrt{\dfrac{1}{9}\cdot [(7-(5))^2+(-4-(5))^2+...+(-2-(5))^2]}\\\\\\[/tex]
[tex]s_1=\sqrt{\dfrac{1}{9}\cdot [(4)+(81)+(169)+(144)+(64)+(100)+(16)+(25)+(36)+(49)]}\\\\\\s_1=\sqrt{\dfrac{688}{9}}=\sqrt{76.44}\\\\\\s_1=8.743[/tex]
The mean and standard deviation for Group 2 is:
[tex]M_2=\dfrac{1}{11}\sum_{i=1}^{11}((-1)+12+(-1)+(-3)+3+(-5)+5+2+(-11)+(-1)+(-3))\\\\\\ M_2=\dfrac{-3}{11}=-0.273[/tex]
[tex]s_2=\sqrt{\dfrac{1}{(n-1)}\sum_{i=1}^{11}(x_i-M)^2}\\\\\\[/tex]
[tex]s_2=\sqrt{\dfrac{1}{10}\cdot [(-1-(-0.273))^2+(12-(-0.273))^2+...+(-3-(-0.273))^2]}\\\\\\[/tex]
[tex]s_2=\sqrt{\dfrac{1}{10}\cdot [(0.53)+(150.62)+...+(115.07)+(0.53)+(7.44)]}\\\\\\s_2=\sqrt{\dfrac{348.182}{10}}=\sqrt{34.82}\\\\\\s_2=5.901[/tex]
a) We have to calculate a 95% confidence interval for the difference between means, with a T-model.
The Group 1, of size n1=10 has a mean of 5 and a standard deviation of 8.743.
The Group 2, of size n2=11 has a mean of -0.273 and a standard deviation of 5.901.
The difference between sample means is Md=5.273.
[tex]M_d=M_1-M_2=5-(-0.273)=5.273[/tex]
The estimated standard error of the difference between means is computed using the formula:
[tex]s_{M_d}=\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}=\sqrt{\dfrac{8.743^2}{10}+\dfrac{5.901^2}{11}}\\\\\\s_{M_d}=\sqrt{7.644+3.166}=\sqrt{10.81}=3.288[/tex]
The degrees of freedom for this test are:
[tex]df=n_1+n_2-1=10+11-2=19[/tex]
The critical t-value for a 95% confidence interval and 19 degrees of freedom is t=1.96.
The margin of error (MOE) can be calculated as:
[tex]MOE=t\cdot s_{M_d}=1.96 \cdot 3.288=6.444[/tex]
Then, the lower and upper bounds of the confidence interval are:
[tex]LL=M_d-t \cdot s_{M_d} = 5.273-6.444=-1.171\\\\UL=M_d+t \cdot s_{M_d} = 5.273+6.444=11.717[/tex]
The 95% confidence interval for the mean is (-1.171, 11.717).
b) This is a hypothesis test for the difference between populations means.
The claim is that increasing the amount of calcium in our diet reduce blood pressure.
Then, the null and alternative hypothesis are:
H_0: \mu_1-\mu_2=0\\\\H_a:\mu_1-\mu_2> 0
The significance level is 0.05.
The sample 1, of size n1=10 has a mean of 5 and a standard deviation of 8.743.
The sample 1, of size n1=11 has a mean of -0.273 and a standard deviation of 5.901.
The difference between sample means is Md=5.273.
[tex]M_d=M_1-M_2=5-(-0.273)=5.273[/tex]
The estimated standard error of the difference between means is computed using the formula:
[tex]s_{M_d}=\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}=\sqrt{\dfrac{8.743^2}{10}+\dfrac{5.901^2}{11}}\\\\\\s_{M_d}=\sqrt{7.644+3.166}=\sqrt{10.81}=3.288[/tex]
Then, we can calculate the t-statistic as:
[tex]t=\dfrac{M_d-(\mu_1-\mu_2)}{s_{M_d}}=\dfrac{5.273-0}{3.288}=\dfrac{5.273}{3.288}=1.604[/tex]
The degrees of freedom for this test are:
[tex]df=n_1+n_2-1=10+11-2=19[/tex]
This test is a right-tailed test, with 19 degrees of freedom and t=1.604, so the P-value for this test is calculated as (using a t-table):
[tex]P-value=P(t>1.604)=0.063[/tex]
As the P-value (0.063) is bigger than the significance level (0.05), the effect is not significant.
The null hypothesis failed to be rejected.
There is not enough evidence to support the claim that increasing the amount of calcium in our diet reduce blood pressure.
