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A survey of 1,500 Canadians reveals that 945 believe that thereis too much sex on television. In a survey of 1,500 Americans, 810believe that there is too much tele...

Question

A survey of 1,500 Canadians reveals that 945 believe that thereis too much sex on television. In a survey of 1,500 Americans, 810believe that there is too much television sex. Can we infer at the99% significance level that the proportion of Canadians andAmericans who believe that there is too much sex on televisiondiffer? Formulate in excel.

A survey of 1,500 Canadians reveals that 945 believe that there is too much sex on television. In a survey of 1,500 Americans, 810 believe that there is too much television sex. Can we infer at the 99% significance level that the proportion of Canadians and Americans who believe that there is too much sex on television differ? Formulate in excel.



Answers

Television Watching The average number of hours of television watched per week by women over age 55 is 48 hours. Men over age 55 watch an average of 43 hours of television per week. Random samples of 40 men and 40 women from a large retirement community yielded the following results. At the 0.01 level of significance, can it be concluded that women watch more television per week than men? (Graph cannot copy)

The following sample data. And we want to say whether at 10 significance, we can conclude the population mean μ does not equal 4.55. Using this data alone. Before we can say anything about the population mean, you we need to complete the sample mean and the sample standard deviation. We compute them as follows The sample mean x bar is some of the data divided by N or 4.76. As we compete on the left and the sample standard deviation us is a square root. The sum of deviations about the mean square divided by N -1. Hear that becomes 2.297. Now, in order to say anything about the population, mean, you were going to have to conduct hypothesis tests for this, me and you, namely, since we have a sample standard deviation s we're going to compute a T test with the following parameters. No hypothesis. H not musicals 4.55 alternative hypothesis. H A new does not equal 4.55 and alpha equals 0.1 for significance. In order to conduct this test, we're gonna have to follow three steps which will go through together one by one. Now, step one of three is compute the T stat of the test statistic. The T status simply given as x minus mu developed. S overruled them. Talking in the data from above gives .41 next. We compute the critical value of the critical region. For a two tailed tests with alpha equals .1, We find in a tea table that for degree of freedom and -1 equals 19 are critical value. TC is plus or -1.7-9. Yeah, That means the critical region is everywhere less than negative, 1.7-9 and everywhere greater than 1.7-9. Finally, we can draw a conclusion 6.41 is not in the critical region I just described. That means that T. Is not in the critical region. That means we cannot reject h not we fail to reject them all hypothesis A. K. We lack evidence that the population we knew does not equal 4.55.

In this problem. We're investigating whether there's a statistically significant difference in the amount of time spent watching television between Children and teens. We are given the results of two samples both of size 15 and we are told that the average number of hours spent viewing television by Children is 22.45 We're given the teens on average, spent 18.5 hours and we're not giving the underlying population Sigma's. So we calculate state a deviation from the samples. Those are going to be 16.4 and 18.2 are no hypothesis is that there's no difference are our alternative hypothesis is merely that they're different, that in fact, there is a justice statistically significant difference between the two averages. Since we're not hypothesizing a specific direction, that is, that either teens air greater than Children or vice versa. This is a two tailed or two sided test. All right, now, our T statistic for this Ah, this is gonna end up being a student t statistic because we have Ah, because we are using the sample standard deviation rather than the Population Sigma. Uh, and also as an assumption, we have to assume that the underlying population is distributed normally because our sample size is less than 30. So that's unnecessary assumption. Or to be able to use the student t distribution. The test statistic that we're going to use is gonna be the difference of the means. So it's gonna be expire for Children minus the expert for teens, minus are hypothesized, expected value and standardized by dividing by the standard error which, with sample standard deviation, is yes, square t over that number as well. Now, since we're hypothesizing that no u sub sea and music tear equal, then we're hypothesizing that this difference is zero and that term goes away. Sorry, test statistic is now adjust this portion here, okay, and that is going to be distributed according to the student T distribution that distribution requires we calculate the degrees of freedom. Degrees of freedom are always equal to the smaller of the two sample sizes minus one In this case, both sample sizes Air 15 we subtract one and that gives us 14 degrees of freedom. We're also told in the problem that we're to do the calculation assuming a 1% or Alfa equals 10.1 level of significance. If we have a student, he distribution with 14 degrees of freedom. It's two sided, and we're looking for an Alfa of Point. A one. This is going to give us a critical value, plus or minus 2.62 So if we were to visualize this, we're looking for distribution. It looks like that with cutoffs, a 2.62 and negative 2.62 If we use the data given and calculate R T statistic using using our formula, we're gonna get a T statistic of 2.601 which, while close, is just that side of the 2.62 Therefore, we do not reject the null hypothesis or said another way. The data does not give us reason to believe that there's a statistically significant difference between the television viewing time of Children versus that of Teague's

Another confidence interval. This time we're gonna be looking at 200 people who watch TV in 154 of them. Watch educational TV. Now we want to get a good 90% estimate for the proportion of people in the general population who watch educational TV. So let's start this time by estimating the proportion from what we have to do. This, we're gonna take 154 the number of people who do watch educational TV divided by the total number of people, and we got 77%. That's R P hat estimate, and we can write this as 0.77 if we want. It's gonna be useful later to note that Q hat is one minus p hat. So that's gonna be 0.23 so that these guys at up to 100% great. Now we can get started and find are expected error. The formula for this is our Z term times a square root of P hat times Q hat all over the total number of people end. Since we're working with a 90% confidence interval, RZ value is 1.65 that you can plug in our values for P and Q, that's 0.77 point 23 divided all by 200 the total number of people. And if we plug that in, we get points 05 or 5% to you prefer percentages. Great. Now we can start actually writing our confidence interval. You know that there's a 90% chance that are true. Value of P is greater than P hat plus 900.5 sorry minus 0.5 and that it's less than P hat plus 0.5. No, it's plug in What we have for P hat and we got P is between 0.72 and 0.82 and that was our 90% confidence interval. Again, it could be useful to write this as just P is equal to 0.77 or estimate plus or minus the standard error 0.5 And these are the two ways that you can write your 90% confidence interval

For this example, we must first calculate the total number of hours in one week. We know that each day has 24 hours and then a week is made up of seven digs. This gives us a total of 168 hours. Now. The report says that an average American spends 31.5 hours watching television. We want to take this amount, and we would divide it by 168 which was our total from the week. This gives us a decimal of 0.1875 We take this decimal and we multiply it times 100 in order to give our percentage, which is 18.75%.


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