A recent study showed that the average number of sticks of gum a person chews in a week is 11. A college student believes that the guys in his dormitory chew less gum in a week. He conducts a study and samples 14 of the guys in his dorm and finds that on average they chew 10 sticks of gum in a week with a standard deviation of 3.1. Test the college student's claim at a=0.01. The correct hypotheses would be: Ο H:μ < 11 HA: µ > 11 (claim) Ο Η: μ 2 11 HA: µ < 11 (claim) Ο Η0: μ 11 HA: µ + 11 (claim) Since the level of significance is 0.01 the critical value is -2.65 The test statistic is: (round to 3 places) The p-value is: (round to 3 places) The decision can be made to: O reject Ho O do not reject Ho The final conclusion is that: O There is enough evidence to reject the claim that less gum in a week. There is not enough evidence to reject the claim that less gum in a week. O There is enough evidence to support the claim that less gum in a week. O There is not enough evidence to support the claim that less gum in a week.
A recent study showed that the average number of sticks of gum a person chews in a week is 11. A college student believes that the guys in his dormitory chew less gum in a week. He conducts a study and samples 14 of the guys in his dorm and finds that on average they chew 10 sticks of gum in a week with a standard deviation of 3.1. Test the college student's claim at a=0.01. The correct hypotheses would be: Ο H:μ < 11 HA: µ > 11 (claim) Ο Η: μ 2 11 HA: µ < 11 (claim) Ο Η0: μ 11 HA: µ + 11 (claim) Since the level of significance is 0.01 the critical value is -2.65 The test statistic is: (round to 3 places) The p-value is: (round to 3 places) The decision can be made to: O reject Ho O do not reject Ho The final conclusion is that: O There is enough evidence to reject the claim that less gum in a week. There is not enough evidence to reject the claim that less gum in a week. O There is enough evidence to support the claim that less gum in a week. O There is not enough evidence to support the claim that less gum in a week.
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 recent study showed that the average number of sticks of gum a person chews in a week is 11. A college student believes that the guys in his dormitory chew less gum in a week. He conducts a study and samples 14 of the guys in his dorm and finds that on average they chew 10 sticks of gum in a week with a standard deviation of 3.1. Test the college student's claim at α=0.01.
The correct hypotheses would be:
- \( H_0: \mu \geq 11 \)
- \( H_A: \mu < 11 \) (claim)
Since the level of significance is 0.01, the critical value is -2.65.
The test statistic is: [_____] (round to 3 places)
The p-value is: [_____] (round to 3 places)
The decision can be made to:
- \( \circ \) reject \( H_0 \)
- \( \circ \) do not reject \( H_0 \)
The final conclusion is that:
- \( \circ \) There is enough evidence to reject the claim that less gum in a week.
- \( \circ \) There is not enough evidence to reject the claim that less gum in a week.
- \( \circ \) There is enough evidence to support the claim that less gum in a week.
- \( \circ \) There is not enough evidence to support the claim that less gum in a week.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd3906bba-d731-4d3e-8323-fa6146946cae%2Fbc72c99a-c82e-4831-b4d1-e9d808f2f288%2Fnb38k6f_processed.png&w=3840&q=75)
Transcribed Image Text:A recent study showed that the average number of sticks of gum a person chews in a week is 11. A college student believes that the guys in his dormitory chew less gum in a week. He conducts a study and samples 14 of the guys in his dorm and finds that on average they chew 10 sticks of gum in a week with a standard deviation of 3.1. Test the college student's claim at α=0.01.
The correct hypotheses would be:
- \( H_0: \mu \geq 11 \)
- \( H_A: \mu < 11 \) (claim)
Since the level of significance is 0.01, the critical value is -2.65.
The test statistic is: [_____] (round to 3 places)
The p-value is: [_____] (round to 3 places)
The decision can be made to:
- \( \circ \) reject \( H_0 \)
- \( \circ \) do not reject \( H_0 \)
The final conclusion is that:
- \( \circ \) There is enough evidence to reject the claim that less gum in a week.
- \( \circ \) There is not enough evidence to reject the claim that less gum in a week.
- \( \circ \) There is enough evidence to support the claim that less gum in a week.
- \( \circ \) There is not enough evidence to support the claim that less gum in a week.
Expert Solution
Step 1
Given information-
Population mean, μ = 11
Sample size, n = 14
Sample mean, M = 10
Sample standard deviation, s = 3.1
Significance level, α = 0.01
Hypothesis Formulation-
Null Hypothesis, H0: μ ≥ 11
Alternate Hypothesis, Ha: μ < 11
Since here population standard deviation is unknown so using t-test statistics
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