Question

In: Math

Biostatistics with R - Solve the following by evaluating the test statistic and p value. Suppose...

Biostatistics with R

- Solve the following by evaluating the test statistic and p value. Suppose that the population mean of systolic blood pressure in the US is 115. We hypothesize mean systolic blood pressure is lower than 115 among people who consume a small amount (e.g., around 3.5 ounces) of dark chocolate every day. Assume that systolic blood pressure, X, in this population has a Normal distribution. To evaluate our hypothesis, we randomly selected 100 people, who include a small amount of dark chocolate in their daily diet, and measured their blood pressure. If the sample mean is x¯ = 111 and the sample variance is s = 32, can we reject the null hypothesis at 0.1 confidence level?

- Solve the following using the Confidence Interval Hypothesis Test approah and t.test() Approach. Use the Pima.tr data set to evaluate the hypothesis that the population mean of diastolic blood pressure for Pima Indian women is not 70.

library(MASS)

data("Pima.tr")

#str(Pima.tr)

?Pima.tr

Solutions

Expert Solution

here the level of significance would be 0.1 not the confidence level. (please check the question once again)

(first part) here we want to test the

null hypothesis H0:=115 and

alternate hypothesis Ha:<115

( this is left one-tailed test, since We hypothesize mean systolic blood pressure is lower than 115 among people who consume a small amount )

here we use t-test and t=|(-)|/(s/sqrt(n))=(111-115)/(32/sqrt(100))=|-1.25|=1.25 with n-1=100-1=99 df

one tailed critical t(0.1,99)=1.66 is more than calcuated t=1.25, so we fail to reject (or accept H0) and conclude that dark chocolate does not lower the systolic blood pressure.

(second part) here we want to test the

null hypothesis H0:=70 and

alternate hypothesis Ha:70(this is two tailed test)

from the given data we found

=71.26 ,s=11.48,n=200

now we can use t-test and

t=|(-)|/(s/sqrt(n))=(71.26-70)/(11.48/sqrt(200))=|1.55|=1.55 with n-1=200-1=199 df

the two tailed critical t(0.1,199)=1.65 is more than calculated t=1.55, so we fail to reject ( or accept )H0 and conclude that diastolic blood pressure for Pima Indian women is 70.


Related Solutions

Conduct the hypothesis test and provide the test​ statistic, critical value, and​ P-value, and state the...
Conduct the hypothesis test and provide the test​ statistic, critical value, and​ P-value, and state the conclusion. A person randomly selected 100 checks and recorded the cents portions of those checks. The table below lists those cents portions categorized according to the indicated values. Use a 0.10 significance level to test the claim that the four categories are equally likely. The person expected that many checks for whole dollar amounts would result in a disproportionately high frequency for the first​...
Conduct the hypothesis test and provide the test​ statistic, critical value and​ P-value, and state the...
Conduct the hypothesis test and provide the test​ statistic, critical value and​ P-value, and state the conclusion. A person drilled a hole in a die and filled it with a lead​ weight, then proceeded to roll it 200 times. Here are the observed frequencies for the outcomes of​ 1, 2,​ 3, 4,​ 5, and​ 6, respectively: 28​, 31​, 44​, 38​, 26​, 33. Use a 0.025 significance level to test the claim that the outcomes are not equally likely. Does it...
Conduct the hypothesis test and provide the test​ statistic, critical value and​ P-value, and state the...
Conduct the hypothesis test and provide the test​ statistic, critical value and​ P-value, and state the conclusion. A person randomly selected 100 checks and recorded the cents portions of those checks. The table below lists those cents portions categorized according to the indicated values. Use a 0.100 significance level to test the claim that the four categories are equally likely. The person expected that many checks for whole dollar amounts would result in a disproportionately high frequency for the first​...
Conduct the hypothesis test and provide the test? statistic, critical value and? P-value, and state the...
Conduct the hypothesis test and provide the test? statistic, critical value and? P-value, and state the conclusion. A person randomly selected 100100 checks and recorded the cents portions of those checks. The table below lists those cents portions categorized according to the indicated values. Use a 0.100.10 significance level to test the claim that the four categories are equally likely. The person expected that many checks for whole dollar amounts would result in a disproportionately high frequency for the first?...
Conduct the hypothesis test and provide the test​ statistic, critical value and​ P-value, and state the...
Conduct the hypothesis test and provide the test​ statistic, critical value and​ P-value, and state the conclusion. A person drilled a hole in a die and filled it with a lead​ weight, then proceeded to roll it 200200 times. Here are the observed frequencies for the outcomes of​ 1, 2,​ 3, 4,​ 5, and​ 6, respectively: 2626​, 3030​, 4848​, 4141​, 2828​, 2727. Use a 0.010.01 significance level to test the claim that the outcomes are not equally likely. Does it...
Conduct the hypothesis test and provide the test​ statistic, critical value and​ P-value, and state the...
Conduct the hypothesis test and provide the test​ statistic, critical value and​ P-value, and state the conclusion. A person randomly selected 100 checks and recorded the cents portions of those checks. The table below lists those cents portions categorized according to the indicated values. Use a 0.10 significance level to test the claim that the four categories are equally likely. The person expected that many checks for whole dollar amounts would result in a disproportionately high frequency for the first​...
Find the P - value for the test statistic z=−1.41 for the following null and alternative...
Find the P - value for the test statistic z=−1.41 for the following null and alternative hypotheses: H0: The population mean is 50.   Ha: The population mean is less than 50.   The P - value is   (b) Find the P - value for the test statistic z=−1.41 for the following null and alternative hypotheses: H0: The population mean is 50.   Ha: The population mean is not equal to 50.   The P - value is
What is the value of the test-statistic? What is the p-value? What is the critical value?
Original Claim: The average age of all patients admitted to the hospital with infectious diseases is less than 65 years of age. Test the claim using ? = 0.05 and assume your data is normally distributed and ? is unknown. Q.) What is the value of the test-statistic? What is the p-value? What is the critical value?   Patient # Infectious Disease Age 1 Yes 67 2 Yes 38 3 Yes 58 4 Yes 52 5 Yes 46 6 Yes...
What would be the p value of the test statistic .0006 for a two sided test...
What would be the p value of the test statistic .0006 for a two sided test and a degree of freedom of 8
1) What is the test statistic? 2) What is the p value for your test?
1) What is the test statistic? 2) What is the p value for your test?
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT