A simple random sample of 20 pages from a dictionary is obtained. The numbers of words defined on those pages are found, with the results n= 20, x= 60.1 words, s= 16.9 words. Given that this dictionary has 1457 pages with defined words, the claim thatthere are more than 70,000 defined words is equivalent to the claim that the mean number of words per page is greater than 48.0 words. Use a 0.01 significance level to test the claim that the mean number of words per page is greater than 48.0 words. Whatdoes the result suggest about the claim that there are more than 70,000 defined words? Identify the null and alternative hypotheses, test statistic, P-value, and state the final conclusion that addresses the original claim. Assume that the population is normallydistributed.....What are the null and alternative hypotheses?O A. Ho: µ>48.0 wordsH1:H<48.0 wordsVB. Ho: H= 48.0 wordsH:p>48.0 wordsO C. Ho: µ = 48.0 wordsO D. Ho: H= 48.0 wordsH,:µ# 48.0 words1<48.0 wordsDetermine the test statistic.3.20 (Round to two decimal places as needed.)Determine the P-value.(Round to three decimal places as needed.)

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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)
Suppose an instructor would like to know if there is difference in the performance of her two courses of 560 students each. She proceeds to take a random sample of size 50 from each independent class and was able to establish the following results: mean SD Class 1 86 17 Class 2 78 15 Is there a significant difference between the two classes at a significance level of five percent?
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
A researcher assures that in his country, on average, the salary of workers is less than $5000.00. He takes a sample of 7 workers and finds that their salary is: $3900.00, $5100.00, $3100.00, $5500.00, $2700.00, $4500.00, $4100.00 Can you be sure at the 0.02 level of significance that the researcher is correct? Choose 1. The correct conclusion2. The correct interpretation Select one:to)1. H0 is rejected2. With a confidence of 98% it can be ensured that the average wage of the workers is not less than $5000.00b)1. H0 is not rejected2. With a confidence of 98% it can be ensured that the average wage of the workers is not less than $5000.00c)1. H0 is not rejected2. It can be ensured, with 98% confidence, that the average wage of the workers is less than $5,000.00d)1. H0 is rejected2. It can be ensured, with 98% confidence, that the average wage of the workers is less than $5,000.00
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.
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 company claims that less than 30% of people who both work and live in Newcastle. A study revealed that among a random sample of 200 people working in Newcastle, 62 also lived in Newcastle. Does the sample support the company’s claim? Use a level of significance of 0.01.
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.
please answer the first 3 and explain the reason for the answer.

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