KA simple random sample of 20 pages from a dictionary is obtained. The numbers of words defined on those pagesare found, with the results n = 20, x=54.3 words, s= 15.2 words. Given that this dictionary has 1436 pages withdefined words, the claim that there are more than 70,000 defined words is equivalent to the claim that the mean numberof words per page is greater than 48.7 words. Use a 0.01 significance level to test the claim that the mean number ofwords per page is greater than 48.7 words. What does the result suggest about the claim that there are morethan 70,000 defined words? Identify the null and alternative hypotheses, test statistic, P-value, and state the final

Assume that μ is the population mean number of words per page.

Answered: A simple random sample of 20 pages from…

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
A Michigan study concerning preference for outdoor activities used a questionnaire with a six-point Likert-type response in which 1 designated "not important" and 6 designated "extremely important." A random sample of n1 = 44 adults were asked about fishing as an outdoor activity. The mean response was x1 = 4.9. Another random sample of n2 = 49 adults were asked about camping as an outdoor activity. For this group, the mean response was x2 = 4.1. From previous studies, it is known that ?1 = 1.5 and ?2 = 2.0. Does this indicate a difference (either way) regarding preference for camping versus preference for fishing as an outdoor activity? Use a 5% level of significance. Note: A Likert scale usually has to do with approval of or agreement with a statement in a questionnaire. For example, respondents are asked to indicate whether they "strongly agree," "agree," "disagree," or "strongly disagree" with the statement.
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)
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Suppose that in a random selection of 100100 colored candies, 2323% of them are blue. The candy company claims that the percentage of blue candies is equal to 2525%. Use a 0.100.10 significance level to test that claim.
A dietitian wishes to see if a person's cholesterol level will change if the diet is supplemented by a certain mineral. Seven subjects were pretested and then took the mineral supplement for a six-week period. The results are shown below in the table. Use a paired samples t-chart at a = 0.01 significance level to see difference (Before - After) between the cholesterol levels. State the hypotheses and conclusion. Also, conduct a 99% confidence interval. Subject 1 Before 210 After 190 2 205 170 3 208 210 4 192 188 5 178 173 6 244 228 7 211 198
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Suppose that in a random selection of 100 colored candies, 23% of them are blue. The candy company claims that the percentage of blue candies is equal to 26%. Use a 0.05 significance level to test that claim.
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Suppose that in a random selection of 100 colored candies, 20% of them are blue. The candy company claims that the percentage of blue candies is equal to 29%. Use a 0.01 significance level to test that claim.
A Michigan study concerning preference for outdoor activities used a questionnaire with a six-point Likert-type response in which 1 designated "not important" and 6 designated "extremely important." A random sample of n1 = 41 adults were asked about fishing as an outdoor activity. The mean response was x1 = 4.9. Another random sample of n2 = 50 adults were asked about camping as an outdoor activity. For this group, the mean response was x2 = 5.7. From previous studies, it is known that ?1 = 1.6 and ?2 = 1.5. Does this indicate a difference (either way) regarding preference for camping versus preference for fishing as an outdoor activity? Use a 5% level of significance.Note: A Likert scale usually has to do with approval of or agreement with a statement in a questionnaire. For example, respondents are asked to indicate whether they "strongly agree," "agree," "disagree," or "strongly disagree" with the statement. (a) What is the level of significance? .05 State the null and alternate…
Suppose that in a random selection of 100 colored candies, 20% of them are blue. The candy company claims that the percentage of blue candies is equal to 22%. Use a 0.01 significance level to test that claim.

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