The 2006 general Social Survey asked a large number of people how much time they spent watching TVeach day. The mean number of hours was 3.35 with a standard deviation of 2.78. Assume that in a sampleof 30 teenagers, the sample standard deviation of daily TV time is 2.37 hours, and that the population ofTV watching times is normally distributed. Under 10% significance level can you conclude that thepopulation standard deviation of TV watching times for teenagers is less than 2.78?Procedure: One variance x² Hypothesis TestAssumptions: (select everything that applies)O Normal populationO Simple random sampleO The number of positive and negative responses are both greater than 10O Sample size is greater than 30OPopulation standard deviation is knownO Population standard deviation is unknownPart 2 of 5Step 1. Hypotheses Set-Up:where ? v is the Select an answerand the units are ? VHo: Select an answer VHa: Select an answerand the test is Select an answer vStep 2. The significance level a =

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Answered: The 2006 general Social Survey asked a…

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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Steve believes that his wife's cell phone battery does not last as long as his cell phone battery. On eleven different occasions, he measured the length of time his cell phone battery lasted and calculated that the mean was 16.4 hours with a standard deviation of 6.4 hours. He measured the length of time his wife's cell phone battery lasted on seven different occasions and calculated a mean of 23.3 hours with a standard deviation of 4.2 hours. Assume that the population variances are the same. Let Population 1 be the battery life of Steve's cell phone and Population 2 be the battery life of his wife's cell phone. Step 1 of 2: Construct a 90 % confidence interval for the true difference in mean battery life between Steve's cell phone and his wife's. Round the endpoints of the interval to one decimal place, if necessary.
Assume the average amount of caffeine consumed daily by adults is normally distributed with a mean of 260 mg and a standard deviation of 48 mg. In a random sample of 600 adults, how many consume at least 320 mg of caffeine daily?
The 2015 general Social Survey asked a large number of people how much time they spent watching TV each day. The mean number of hours was 4.2 with a standard deviation of 2.8. Assume that in a sample of 39 teenagers, the sample standard deviation of daily TV time is 2.8 hours, and that the population of TV watching times is normally distributed. Under 10% significance level can you conclude that the population standard deviation of TV watching times for teenagers is less than 2.8? Procedure: Select an answer Assumptions: (select everything that applies) Simple random sample Population standard deviation is unknown O Sample size is greater than 30 Normal population Population standard deviation is known The number of positive and negative responses are both greater than 10
Average adult Americans are about one inch taller but nearly a whopping 25 pounds heavier than they were in 1960, according to a report from the Centers for Disease Control and Prevention (CDC). City A is considered one of America's healthiest cities. Is the weight gain since 1960 similar in city A? A sample of n=25 adults suggested a mean increase of 22 pounds with a standard deviation of 9.4 pounds. Is City A significantly different in terms of weight gain since 1960? Complete and interpret 5 step hypothesis testing by hand at a 5% level of significance using a Z-test. Show each step and your work.
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A marketing researcher wants to estimate the mean amount spent per year ($) on a web site by membership member shoppers. Suppose a random sample of 100 membership member shoppers who recently made a purchase on the web site yielded a mean amount spent of $58and a standard deviation of $55.Complete parts (a) and (b) below. a. Is there evidence that the population mean amount spent per year on the web site by membership member shoppers is different from $51? (Use a 0.01 level of significance.) State the null and alternative hypotheses.
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In the class of 2019, more than 1.6 million students took the SAT. The distribution of scores on the math section (out of 800) is approximately normal with a mean of 528 and standard deviation of 117. What proportion of students earned scores between 500 and 600 on the SAT math test?
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Steve believes that his wife's cell phone battery does not last as long as his cell phone battery. On eight different occasions, he measured the length of time his cell phone battery lasted and calculated that the mean was 19.7 hours with a standard deviation of 9.4 hours. He measured the length of time his wife's cell phone battery lasted on twelve different occasions and calculated a mean of 18.9 hours with a standard deviation of 5.3 hours. Assume that the population variances are the same. Let Population 1 be the battery life of Steve's cell phone and Population 2 be the battery life of his wife's cell phone. Step 1 of 2: Construct a 90 % confidence interval for the true difference in mean battery life between Steve's cell phone and his wife's. Round the endpoints of the interval to one decimal place, if necessary.

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