Ages of Gamblers The mean age of a sample of 25 people who were playing the slot machines is 52 years, and the standard deviation is 6.8 years. The meanage of a sample of 35 people who were playing roulette is 53 with a standard deviation of 3.2 years. Can it be concluded at a=0.01 that the mean age of thoseplaying the slot machines is less than those playing roulette? Use μ, for the mean age of those playing slot machines. Assume the variables are normallydistributed and the variances are unequal. Use a TI-83 Plus/TI-84 Plus calculator.Part 1 of 4(a) State the hypotheses and identify the claim with the correct hypothesis.Ho: H = H2 not claimH₁ H₁ H₂This hypothesis test is a two-tailed ▼test.Part 2 of 4claimP-value=XO<0 0>0 O*O30=0 441(b) Find the P-value. Round the answer to at least four decimal places.X3H₂ Part: 2/4Part 3 of 4(c) Make the decision.(Choose one) the null hypothesis.Part: 3/4Part 4 of 4S(d) Summarize the results.There (Choose one) enough evidence to support the claim that the mean age of those playing the slot machines is lessthan those playing roulette.X

For the given data we need to test the two sample t test for unequal variances.

Answered: Ages of Gamblers The mean age of a…

A First Course in Probability (10th Edition)

10th Edition

ISBN:9780134753119

Author:Sheldon Ross

Publisher:Sheldon Ross

Chapter1: Combinatorial Analysis

Section: Chapter Questions

Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...

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A community college Math instructor feels that the average age of community college students decreased since they started teaching 25 years ago. Looking up data from the year the instructor started teaching, the mean age of community college students was 24.3 years old, with a population standard deviation of 2.2 years. The Math instructor took a sample of 40 of their current community college students and found that the sample had a mean age of 23.7 years old. Assume the population of ages of community college students is normally distributed and that the population standard deviation has remained the same. Test at the 5% significance level if the mean age of community college students is now less than 24.3 years old. Choose the correct answer from the parenthesis. (i) Hypotheses ?0: ? = ( 0, 2.2, 22.7, 24.3, 25, 40) ______________ . ?1: ? = ( < , > ) ______________ ( 0, 2.2, 22.7, 24.3, 25, 40) ________________ . (ii) Test Statistic ( z/ t )…
Use z scores to compare the given values. In a recent awards ceremony, the age of the winner for best actor was 34 and the age of the winner for best actress was 43. For all best actors, the mean age is 42.5 years and the standard deviation is 6.2 years. For all best actresses, the mean age is 31.9 years and the standard deviation is 11.2 years. (All ages are determined at the time of the awards ceremony.) Relative to their genders, who had the more extreme age when winning the award, the actor or the actress? Explain. Question content area bottom Part 1 Since the z score for the actor is z=enter your response here and the z score for the actress is z=enter your response here, the ▼ actress actor had the more extreme age. (Round to two decimal places.)
Use z scores to compare the given values. In a recent awards ceremony, the age of the winner for best actor was 33 and the age of the winner for best actress was 47. For all best actors, the mean age is 44.4 years and the standard deviation is 6.5 years. For all best actresses, the mean age is 32.8 years and the standard deviation is 11.2 years. (All ages are determined at the time of the awards ceremony.) Relative to their genders, who had the more extreme age when winning the award, the actor or the actress? Explain. Since the z score for the actor is z= and the z score for the actress is z = the V had the more extreme age. (Round to two decimal places.)
Time to Complete Final Exam. The time needed to complete a final examination in a particular college course is normally distributed with a mean of 80 minutes and a standard deviation of 10 minutes. Answer the following questions. Assume that 20% of students did not complete the exam. What was the exam duration?
The amount of snowfall falling in a certain mountain range is normally distributed with a mean of 80 inches,, and a standard deviation of 10 inches. What is the probability that the men annual snowfall during 25 randomly picked years will exceed 82.8 inches? Draw a picture. Show z calculation. Use Table A-2.
For a certain type of job, it costs a copany an average of $231 to train an employee to perform the task. The standard deviation is $5. Find the minimum percentage of data values that will fall in the range of $219 to $243. Use Chebyshev's theorem.
Scores on a recent NYS Standardized Exam were normally distributed with a mean of 77.6 and a standard deviation of 4.3. Find the NYS Standardized Exam score that separates the bottom 2.5% from the rest. Round this score to the nearest tenth
The income of trainess at a local mill are normally distributed with a mean of $1100 and a standard deviation of $150. What percentage of trainees earn less than $900 a month?
Use z-scores to make the following comparison. A high school student took two college entrance exams, scoring 1110 on the SAT and 27 on the ACT. Suppose that SAT scores have a mean of 950 and a standard deviation of 130 while the ACT scores have a mean of 22 and a standard deviation of 4. Assuming the performance on both tests follows a normal distribution, determine which test the student did better on.
You have a sample of 48 randomly selected Vancouverites. On average (the sample mean), the people in your sample spent 425 hours eating food in 2020. The least time a person in the sample spent eating food in 2020 is 367 hours and the most time a person in the sample spent eating food in 2020 is 484 hours. What would be your best guess about the population standard deviation of the time a Vancouverite spent eating food in 2020? Give the number and explain (including assumptions you make about the population distribution).
Use z scores to compare the given values. In a recent awards ceremony, the age of the winner for best actor was 28 and the age of the winner for best actress was 44. For all best actors, the mean age is 42.3 years and the standard deviation is 8.4 years. For all best actresses, the mean age is 33.3 years and the standard deviation is 10.7 years. (All ages are determined at the time of the awards ceremony.) Relative to their genders, who had the more extreme age when winning the award, the actor or the actress? Explain. Since the z score for the actor is z = and the z score for the actress is z=, the had the more extreme age. (Round to two decimal places.)
A community college Math instructor feels that the average age of community college students decreased since they started teaching 25 years ago. Looking up data from the year the instructor started teaching, the mean age of community college students was 24.3 years old, with a population standard deviation of 2.2 years. The Math instructor took a sample of 40 of their current community college students and found that the sample had a mean age of 23.7 years old. Assume the population of ages of community college students is normally distributed and that the population standard deviation has remained the same. Test at the 5% significance level if the mean age of community college students is now less than 24.3 years old. The question is attached.

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