Assume that the paired sample data are simple random samples...

Assume that the paired sample data are simple random samples and that the differences have a distribution that is approximately normal. Self-Reported and Measured Male Heights As part of the National Health and Nutrition Examination Survey, the Department of Health and Human Services obtained self reported heights and measured heights for males aged $12-16 .$ All measurement are in inches. Listed below are sample results. a. Is there sufficient evidence to support the claim that there is a difference between self reported heights and measured heights of males aged $12-16 ?$ Use a 0.05 significance level. b. Construct a $95 \%$ confidence interval estimate of the mean difference between reported heights and measured heights. Interpret the resulting confidence interval, and comment on the implications of whether the confidence interval limits contain 0. Reported height $\quad\begin{array}{cccccccccccc}68 & 71 & 63 & 70 & 71 & 60 & 65 & 64 & 54 & 63 & 66 & 72\end{array}$ Measured height $\quad\begin{array}{cccccccccccc}67.9&69 .9&64 .9&68.3&70.3&60.6&64.5&67.0&55.6&74.2&65.0&70.8\end{array}$

Transcript

00:01 So this question of us give us some data and we are asked to perform an hypothesis test on the claim that there is actually to support the claim that there's a difference between self -reported self -reported height and measured height of meal between the age of 12 to 16.

00:17 We are also asked to construct a confidence interval for the difference in population mean.

00:24 And so one of the first step is to get, so one of the first step in solving this question is to get the, value of the sample mean and the value of the sample various of our data set so so we have the self -reported heights to be here we have self -reported heights to be here and there we have measured heights to be here so this is a data set for self -reported height i'm using a calculator to get the value of the sample mean and the sample variance so i click on calculate our sample mean is 65 .58 so that means i can see x1 bar is 61 points 65 .58 rather.

01:05 I can call this my sample one and i call the measured height my sample two.

01:14 So our variance, our variance is 27 .90.

01:22 So we have s1 squared to be 27 .90.

01:31 And we have our n1 as 12.

01:33 For our second data, so this is the data for the measured height and then we click calculate.

01:38 As you can see, our mean is 66 .58 so that means our x2 bar is 66 .58 the n2 is also 12 and the variance of the second sample is equal to we have our variance as 24 .67 so our variance is 24 .67 so our variance is 24 .67 so the first step is to state our null and alternative hypothesis and it states that h not states that x1 bar is because to x2 by meaning there is no difference between the self -reported height and the measured height and the alternative hypothesis is stating that x1 bar is not equals to x2 but meaning there is actually a difference now our test statistics is going to be a t -test and this is because our sample sizes is actually lesser than 30 that is we can see 12 here we can see two of here they are both lesser than 30 and also we are not making use of our population standard division we are rather making use of the sample standard deviation of the sample variance so our test statistics is going to be a t test and this is given below and this is given below as t is equals to x1 bar minus x2 bar divided by the square root of s1 squared that is the variance of the first sample divided by n1 and the variance of the second sample divided by n2 so it's time to substitute our parameters into the formula so we'll be having 65 .58 minus 66 .58 divided by the square root of we have 27 .90 divided by 12 plus 24 .67 divided by 12.

03:25 So 65 .58 minus 66 .58 that gives us minus 1 divided by.

03:32 So let us use our calculator to get everything and the denominator we have 27 .90 divided by 12 plus 24 .69 divided by 12 and that gives us 2 .09344.

03:49 So when we divide this we have 1 divided by that we have our test statistics as minus 0 .478.

03:58 So our test statistics is minus 0 .478.

03:59 So our test statistics is minus 0 .478.

04:04 So the next step is for us to get the p value...

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