8. During a one sample t-test the researcher was interested...

8. During a one sample t-test the researcher was interested in finding out the critical value for his test. If the researcher had a sample of 100 subjects then find the critical value from the table below: df 0.10 0.05 0.01 97 28.41 31.41 37.57 98 30.81 33.92 40.29 99 32.01 35.17 41.64 100 33.20 36.42 42.98 101 34.38 37.65 44.31 For this test the critical value would be equal to? 9. Choose the correct answer. When correlation is found between two variables then: a. This means that the relationship is causative b. The relationship is not necessarily available at the individual level c. The relationship might be caused by other variables d. Choices b and c are correct 10. Choose the correct answer. If the sample size is large enough then the value of the standard deviation of the sampling distribution (standard error) is: a. Larger than the population standard deviation b. smaller than the population standard deviation c. Equals the population standard deviation d. Choices a and b are correct

Transcript

00:01 Now here we look at a few little questions, right? the first one says, a researcher wishes to estimate the number of a household is with two computers, right? but our example is needed in order to be 99 % confident that the sample proportion were not to fall from the true proportion by more than 3%.

00:18 But previous that in case the proportion has to with two computers is 20%.

00:21 So basically the margin of error is 3%.

00:23 So let's suppose the margin area is 3%.

00:25 And this must be given by the z score, right, at the 99 % confidence level.

00:34 That translates in the z score at actually 1 minus 99 % divided by 2, so it's 0 .5 % level.

00:42 And then, of course, this has to multiply by the standard revision, which actually is given, is going to be given by 20 % times 1 minus 20%, it's as 80 % and divided by the sample size, right? and then take squared load.

00:54 So the sample size n is something we need to find out.

00:59 Now you can find it out by square this expression, right? and you'll find this to be n equals just z .5 % divided by 3 % and square it and times 20 % times 80 % that's obviously it's given by, you know, it's just 1 .6, 0 .16, right? so 0 .16, right? right.

01:26 So because it's 0 .2 times 0 .8, it's 0 .16.

01:30 And of course, you need to find this score, right? this score i found from the z table, and i found to be actually given by 2 .57.

01:40 Okay? and then you found this to be given by 1 .175.

01:47 Okay, so that's the answer.

01:53 So the correct choice actually is b, right? determine the point estimate of the population mean and found this to be approximately 1 .1 .80.

02:06 Okay.

02:07 So the correct answer is 1 .180.

02:11 Disseman the point estimated of the population mean a margin error of the confidence integral with lower bond and upbound.

02:16 There is right.

02:21 So it's very clear that this value, 17 and 20, if you do the difference, if you take the difference, you get 10, right? it divided by 2, you get a margin error, right? so the correct answer can only be b, right? because only b has the margin of error being five, right? the margin of error has to, if you double the margin of error, you should get the lens of the convenience interval, right? that's 10, right? now, construct a 95 % confusing interval for the population mean, assume the population's normal distribution, and a sample 20, as part of workers, right, has an annual earnings of this much and standardization of this much, right? so what would be the confusing interval? of course, it has two endpoints, right? so for this purpose, let me first work out the margin of error, right? i'm going to call it e, right? the margin of error was given by the z score at this 95 % level.

03:16 So that actually is this squared at 2 .5%.

03:18 And then times the standard deviation, which actually is 6t, 7, 7, divided by the square root of the sample size, which is 20, right? and then you will find this to be given by.

03:29 Of course, you need to find the z score here.

03:32 Under this score, you can find from the table, which i have on this to be.

03:35 Actually 1 .96...

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