On the basis of a physical examination, a doctor determines the probability of no tumour   (event labelled C for ‘clear’), a benign tumour (B) or a malignant tumour (M) as 0.7, 0.2 and 0.1 respectively.

A further, in depth, test is conducted on the patient which can yield either a negative (N) result or positive (P). The test gives a negative result with probability 0.9 if no tumour is present (i.e. P(N|C) = 0.9). The test gives a negative result with probability 0.8 if there is a benign tumour and 0.2 if there is a malignant tumour.

(i) Given this information calculate the joint and marginal probabilities and display in the table below.

2. Obtain the posterior probability distribution for the patient when the test result is

            a) positive,   b) negative

4. Comment on how the test results change the doctor’s view of the presence of a tumour.

The article “Application of Surgical Navigation to To-
tal Hip Arthroplasty” (T. Ecker and S. Murphy, Journal

of Engineering in Medicine, 2007:699–712) reports
that in a sample of 113 people undergoing a certain
type of hip replacement surgery on one hip, 65 of them
had surgery on their right hip. Can you conclude that
frequency of this type of surgery differs between right
and left hips?

(a) State the null and alternative hypotheses.

(b) Use a two-sided 95% Agresti confidence interval for the population proportion to conduct this hypoth-

esis test. After constructing your confidence interval, state the conclusion in the context of the study

and the conclusion of the hypothesis test.

(c) Go ahead and formally conduct the hypothesis test using a = 0.05. Calculate the test statistic and

P-value appropriately. Does your conclusion to the hypothesis test agree with part (b)?

3. It is thought that the mean length of trout in lakes in a certain region is 20

Inches. A sample of 46 trout from one particular lake had a sample mean of 18.5 inches and a sample standard deviation of 4 inches. Conduct a hypothesis test at the 0.05 level to see if the average trout length in this lake is less than mu=20 inches.

The college bookstore tells prospective students that the average cost of its textbooks is $52 with a standard deviation of $4.50. A group of smart statistics students thinks that the average cost is higher. In order to test the bookstore’s claim against their alternative, the students will select a random sample of size 100. Assume that the mean from their random sample is $52.80. Test at 10% significance level.

--> Perform a hypothesis test and state your decision.

For this problem, carry at least four digits after the decimal in your calculations. Answers may vary slightly due to rounding.

In a random sample of 69 professional actors, it was found that 39 were extroverts.

(a) Let p represent the proportion of all actors who are extroverts. Find a point estimate for p. (Round your answer to four decimal places.)


(b) Find a 95% confidence interval for p. (Round your answers to two decimal places.)

lower limit:

upper limit:

1. Find the mean and the Variance of the following sample data:

Discuss the following in 175 words: The concept of mean & standard deviations of probability distributions play a significant role in managerial decision-making.

A random sample of 200 cars has 18 that are green in color. What is the confidence interval for the true proportion of green cars that lie within a 95% confidence interval?

Discuss the following in 175 words: How normal distributions play a significant role in managerial decision-making.

High rent district:the mean monthly rent for a one-bedroom apartment without a doorman in Manhattan is 2544.Assume the standard deviation is 483.A real estate firm samples 83 apartments.Use cumulative distribution table if needed.

What is the probability that the sample mean rent is greater than 2614? Round atleast 4 places

What is the probability that sample mean rent is between 2413 and 2513? Round atleast 4 places

Find the 65th percentile of the sample mean rent? Round 4 places

Would it be unusual if the sample mean were greater than 2610? Round 4 places

Let's focus on the relationship between the average debt in dollars at graduation (AveDebt) and the in-state cost per year after need-based aid (InCostAid).

a) Does a linear relationship between InCost Aid and AveDebt seem reasonable? Explain.

b) Are there any unusual cases in this sample? If yes, state which ones they are and how they may be affecting the least-squares model fit.

Let x = age in years of a rural Quebec woman at the time of her first marriage. In the year 1941, the population variance of x was approximately σ² = 5.1. Suppose a recent study of age at first marriage for a random sample of 51 women in rural Quebec gave a sample variance s² = 2.4. Use a 5% level of significance to test the claim that the current variance is less than 5.1. Find a 90% confidence interval for the population variance.

(a) What is the level of significance?


State the null and alternate hypotheses.

H_o: σ² = 5.1; H₁: σ² < 5.1

H_o: σ² = 5.1; H₁: σ² ≠ 5.1    

H_o: σ² < 5.1; H₁: σ² = 5.1

H_o: σ² = 5.1; H₁: σ² > 5.1

(b) Find the value of the chi-square statistic for the sample. (Round your answer to two decimal places.)


What are the degrees of freedom?


What assumptions are you making about the original distribution?

We assume a normal population distribution.

We assume a binomial population distribution.    

We assume a uniform population distribution.

We assume a exponential population distribution.

(c) Find or estimate the P-value of the sample test statistic.

P-value > 0.100 0.050 < P-value < 0.100    

0.025 < P-value < 0.050

0.010 < P-value < 0.025

0.005 < P-value < 0.010

P-value < 0.005

(d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis?

Since the P-value > α, we fail to reject the null hypothesis.

Since the P-value > α, we reject the null hypothesis.    

Since the P-value ≤ α, we reject the null hypothesis.

Since the P-value ≤ α, we fail to reject the null hypothesis.

(e) Interpret your conclusion in the context of the application.

At the 5% level of significance, there is insufficient evidence to conclude that the variance of age at first marriage is less than 5.1.

At the 5% level of significance, there is sufficient evidence to conclude that the variance of age at first marriage is less than 5.1.    

(f) Find the requested confidence interval for the population variance. (Round your answers to two decimal places.)

Interpret the results in the context of the application.

We are 90% confident that σ² lies outside this interval.

We are 90% confident that σ² lies within this interval.    

We are 90% confident that σ² lies above this interval.

We are 90% confident that σ² lies below this interval.

Tire lifetime:the lifetime of a certain type of automobile tire(in thousands of moles) is normally distributed with mean u=41 and standard deviation o=4 What is the probability that a radon chosen tire has lifetime greater than 50 thousand miles? What proportion of tires have lifetime between 36 and 45 thousand miles What proportion of tires have lifetimes less than 46 thousand miles? Round answers at least 4 places

The age distribution of the Canadian population and the age distribution of a random sample of 455 residents in the Indian community of a village are shown below.

Use a 5% level of significance to test the claim that the age distribution of the general Canadian population fits the age distribution of the residents of Red Lake Village.

(a) What is the level of significance?


State the null and alternate hypotheses.

H₀: The distributions are the same.
H₁: The distributions are different.

H₀: The distributions are different.
H₁: The distributions are the same.    

H₀: The distributions are different.
H₁: The distributions are different.

H₀: The distributions are the same.
H₁: The distributions are the same.

(b) Find the value of the chi-square statistic for the sample. (Round your answer to three decimal places.)


Are all the expected frequencies greater than 5?

Yes No    

What sampling distribution will you use?

chi-square

binomial    

uniform

normal

Student's t

What are the degrees of freedom?


(c) Estimate the P-value of the sample test statistic.

P-value > 0.100 0.050 < P-value < 0.100    

0.025 < P-value < 0.050

0.010 < P-value < 0.025

0.005 < P-value < 0.010

P-value < 0.005

(d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis that the population fits the specified distribution of categories?

Since the P-value > α, we fail to reject the null hypothesis.

Since the P-value > α, we reject the null hypothesis.   

Since the P-value ≤ α, we reject the null hypothesis.

Since the P-value ≤ α, we fail to reject the null hypothesis.

(e) Interpret your conclusion in the context of the application.

At the 5% level of significance, the evidence is insufficient to conclude that the village population does not fit the general Canadian population.

At the 5% level of significance, the evidence is sufficient to conclude that the village population does not fit the general Canadian population.