Question

3) Consider the following exercise: Suppose that a student is taking a multiple-choice exam in which each question has four choices. Assuming that she has no knowledge of the correct answer to any of the questions, she has decided on a strategy in which she will place four balls (marked A, B, C, and D) into a box. She randomly selects one ball for each question and replaces the ball in the box. The marking on the ball will determine her answer to the question. What is the probability she will get at least four questions correct.

6) A consulting engineering company claims that its employee' cell phone bills average less than $100 per month. A random sample of 75 employees reported an average monthly billing of $94.25 per month with a standard deviation of $17.38. Test the claim at the α = 0.05 level of significance.

7) A heavy equipment liability insurance company claims that the average bulldozer in use is less than six years old. A random sample of fifteen bulldozers had an average age of 5.4 years with a sample deviation of 1.1 years. Assume the population is normally distributed. Test the claim at the α = 0.05 level of significance.

Answer 1

(1) For This, we need the total number of questions in the exam

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(2) Given: = $100 / Month, = $94.25 / Month, s = $17.38 / Month, n = 75, = 0.05

The Hypothesis:

H0: = 100: The average cell phone bill is equal to $100.

Ha: < 100: The average cell phone bill is lesser than $100. (Claim)

This is a Left tailed test

The Test Statistic: Since the population standard deviation is unknown, we use the students t test..

The test statistic is given by the equation:

t observed = -2.87

The p Value: The p value (Left tailed) for t = -2.87, for degrees of freedom (df) = n-1 = 74, is; p value = 0.0027

The Critical Value: The critical value (Left Tail) at = 0.05, for df = 74, tcritical= -1.666

The Decision Rule:  

The Critical Value Method: If tobserved is < -tcritical.

The p-value Method: If P value is < , Then Reject H0.

The Decision:

The Critical Value Method: Since tobserved (-2.87) is < -t critical (-1.666), we Reject H0.

The p-value Method: Since P value (0.0012) is < (0.05) , We Reject H0.

The Conclusion: There is sufficient evidence at the 95% significance level to conclude that the average cell phone bill is lesser than $100.

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(3) Given: = 6 years, = 5.4 years, s = 1.1 years, n = 15, = 0.05

The Hypothesis:

H0: = 100: The average bulldozer in use is equal to 6 years.

Ha: < 100: The average bulldozer in use is lesser than 6 years. (Claim)

This is a Left tailed test

The Test Statistic: Since the population standard deviation is unknown, we use the students t test..

The test statistic is given by the equation:

t observed = -2.11

The p Value: The p value (Left tailed) for t = -2.11, for degrees of freedom (df) = n-1 = 14, is; p value = 0.0267

The Critical Value: The critical value (Left Tail) at = 0.05, for df = 14, tcritical= -1.761

The Decision Rule:  

The Critical Value Method: If tobserved is < -tcritical.

The p-value Method: If P value is < , Then Reject H0.

The Decision:

The Critical Value Method: Since tobserved (-2.11) is < -t critical (-1.761), we Reject H0.

The p-value Method: Since P value (0.0267) is < (0.05) , We Reject H0.

The Conclusion: There is sufficient evidence at the 95% significance level to conclude that the average bulldozer in use is lesser than 6 years.

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