In 2015, the average duration of long-distance telephone calls from a certain town was 9.4minutes. A telephone company wants to perform a test to determine whether this averageduration of long-distance calls has changed. Twenty calls, originating from the town, wasrandomly selected and tne mean duration was 10.2 minures with standard deviation 4.8minutes.a) Using a 1% level of significance, Complete the following:iv. State the decision rule.v. Give your decision based on the available sample evidence.vi. Hence, state your conclusion.

Null Hypothesis: A hypothesis which is tested for plausible rejection is called the Null Hypothesis…

Answered: In 2015, the average duration of…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Previously, an organization reported that teenagers spend 8.5 hours per week, on average, on the phone. The organization thinks that, currently, the mean is higher. Twenty-three randomly chosen teenagers were asked how many hours per week they spend on the phone, with the following results (in hours): 8.3, 8.5, 10.5, 2.9, 6.7, 7.7, 13.8, 7.8, 8.7, 11.8, 12, 6.1, 5.4, 5.8, 8.5, 7.9, 10.1,11.1, 11.6, 12.9, 3.6, 6.4, 11.2 Perform a hypothesis test using a 3% level of significance. Step 1: State the null and alternative hypotheses. Ho: ??v H: ? v ? v (So we will be performing a Select an answer v test.) Step 2: Assuming the null hypothesis is true, determine the features of the distribution of point estimates using the Central Limit Theorem. v with By the Central Limit Theorem, we know that the point estimates are Select an answer and distribution standard deviation distribution mean Step 3: Find the p-value of the point estimate. P? v? v = P(| ? ♥| ? v
The freethrow percentages for the regional basketball tournament participants were compiled. The percents ranged from 32% to 93% with a mean of 51% and a standard deviation of 7%. The quartiles and median percentages were 42%, 81%, and 51%. Is the distribution symmetric, skewed to the left, or skewed to the right? Choose the best answer below: A. Skewed to the right, mean higher than median. B. Symmetric, mean same as median. C. Skewed to the left, mean higher than median. D. Skewed to the left, mean lower than median. E. Skewed to the right, mean lower than median.
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In 2015, the average duration of long-distance telephone calls from a certain town was 3.9 minutes.A telephone company wants to perform a test to determine whether this average duration of longdistance calls has changed. Fifty calls, originating from the town, was randomly selected and thefollowing summary minutes.∑ ? = 205 ∑(? − ?̅)∑(? − ?̅)2 = 56.43 a) Calculate the sample mean, ?̅. b) Calculate the sample standard deviation c) Using the p value approach at the 1% level of significance, test whether the mean duration oflong-distance calls from the town had increased. d) Construct the 99% confidence interval for the population mean duration of the long-distance calls from the town. Answer sub-parts (d)
make up 2 data sets, each with 5 values, that have the same mean and median, but different standard deviations.
A study was conducted on the average commuting distance of students at College. The study found that the first sample of thirty students has an average commute of 23.1 miles with a sample standard deviation of 12.47 miles. The commuting mileages for a second sample of 30 students are below. a) Find the mean, median and mode for the second sample of students. b) Find the range, variance and standard deviation for the second sample students. c)
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A researcher wanted to find the best viewing distance for watching a TV screen. He recruited the following number of participants and asked them to evaluate their satisfaction of the viewing distance. Then, he conducted an ANOVA to see if there is any significant mean differences. Here are the number of participants by each group: Group 1 (4 feet) - 5 people Group 2 (7 feet) - 5 people Group 3 (10 feet) - 6 people Group 4 (13 feet) - 5 people Group 5 (16 feet) - 4 people Given the numbers as presented above, what is the degrees of freedom for within groups for this analysis? a. 4 b. 20 O c. 21 d. 25

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