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A social researcher claims that the average adult Iistens to the radio less than 26 hours per week: He collects data on 25 individuals' radio listening habits ...

Question

A social researcher claims that the average adult Iistens to the radio less than 26 hours per week: He collects data on 25 individuals' radio listening habits and finds that the mean number of hours that the 25 people spent listening to the radio was 22.4 hours: If the population standard deviation is known t0 be eight hours, what is the p-value associated with the test statistic?Select one:0.02440.01220.00990.03640.0198

A social researcher claims that the average adult Iistens to the radio less than 26 hours per week: He collects data on 25 individuals' radio listening habits and finds that the mean number of hours that the 25 people spent listening to the radio was 22.4 hours: If the population standard deviation is known t0 be eight hours, what is the p-value associated with the test statistic? Select one: 0.0244 0.0122 0.0099 0.0364 0.0198



Answers

One hundred eight Americans were surveyed to determine the number of hours they spend watching television each month. It was revealed that they watch an average of 151 hours each month, with a standard deviation of 32 hours. Assume that the underlying population distribution is normal. Define the random variable $\overline{x}=$ in words.

So this problem wants us to find the Nolan alternative hypothesis to this scenario. Essentially, you're working of means. I want to make one housekeeping comment burst. Normally, in statistics, you write mean as the letter mule. That is what I am going to be doing now. No, an alternative hypothesis. We write them like this. H not. Is are no hypothesis, and H. A is our alternative hypothesis. So how do we tell which one is which? Well, the alternative hypothesis is usually what the researchers trying to test for. It could also mean that there's change between two groups, and it is usually the wine that doesn't have an equal to option. So in this problem, they now think that the meanest higher this is the key phrase because it tells us that the alternative hypothesis is going to be mule are mean, is now higher or greater than now. What was the original number 4.5? Now the null hypothesis. There's two ways you could think about this. If the alternative hypothesis is that the change in mean is getting higher, well, that could mean that the null hypothesis is that it is not getting higher mu is less than or equal to 4.5. However, if you think about this, as the alternative hypothesis is, the mean is changing from 4.5, then it could be that the null hypothesis is mu is exactly equal to 4.5. Both of these would give you the same answer if we were to actually do this whole problem out. However, this is a multiple choice question, so you're just gonna have to go with whichever one matches the answer options, and we are done.

Problem. 21. We have a report in 2010 that indicates that Americans between the ages, the ages of the Americans between eight and 18 years it's been an average of new equals 7.5 hours per day using some sort of training device, and in this report, the Standard Division was it will 2.5 hours per day. It's required to find the probability of selecting an individual who uses electronic devices more than nine hours a day. Then for about A it's required to get the probability of X is greater than nine hours birthday. We can get this probability using the school instead. The score equals X minus mu, divided by sigma we have X here equals nine minus mu, which is given 7.5 Divided by sigma equals 2.5. Then the score for nine hours per day equals 4.6. By entering the table, we can get the probability that zit is greater than zero and smaller than 2.6. The tables we have gives the probability starting from zero to the specific is it which is also in six gives in this area the probability from the table equals four point oh 793 then took it. The probability that is greater than 4.6 or greater than X equals nine. Then the probability of X is greater than nine equals half, which is half the area of the normal distribution buying us the shaded area. Sorry, this value is not opened or 73 equals point two 57 Then the probability is half minus 0.2257 which equals 4.2743 which equals 27.43%. Then the probability to select an individual who uses electronic devices more than nine hours per day equals 27.43%. For Part B. We want to get the portion of the same age from 8 to 18 year old who spent between eight and 12 hours. She means we need to get the probability for Ex is greater than eight hours and smaller than 12 hours. Birthday, of course. Then, to answer this, we get to the scores the first season score for it. We use X as 88 hours per day, eight minus 7.5, divided by 2.5 gives open to and for the two we get X equals 12, then 12 minus 7.5. Divided by 2.5, it equals 1.8. Then we get the probability for each Z score from the tables for from zero that is greater than zero, said one. And between 1.6 Sorry. Here it's open to from the tables. It equals four point oh 793 And for the second school between 1.8, it equals 4.4 641 This means we have the normal distribution here is open to and here is one point. The first area here is for that equal 0.2. It's the probability is this value and we have another venue from zero to 1.8. We have poll this area. Then to get the area to get only this area, we'll get the probability for X. Between eight and 12 hours per day will be the shaded area with blue, which is opened 4641 minus the shaded area with black Oh, point Oh 793 then equals Oh, 0.38 for it or we can write it in percentage 38.48% and this is the final answer of body and this is the final answer of 40 and the problem.

Hi. This question. We got some information about the teeth, this thing and we get that T is equal to minus 1.9 on Devalue off 1139 On the degree of freedom is anyone So as we know that the great freedom is equal Toe end minus one. So the assistant for end is a photo. Your freedom plus one just of anyone. This one So the business summit is

46 which distribution should use for this problem in this case were given an average of 151 and we're in a standard deviation that's representative of the sample instead of the population. Because we're given this, the distribution we should use is a T distribution.


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