Q3: The heights of a random sample of a particular species...

Q3: The heights of a random sample of a particular species of horse were measured (in cm) as: 45.3 47.1 44.2 46.8 46.5 45.5 47.6 Assume that heights are normally distributed, (a) Find a point estimate for the mean height, and its standard error. (b) Find the maximal margin of error with 95% confidence level. (c) compute a 95% confidence interval for the mean. Interpret your results. (d) Is there evidence that the mean height is higher than 45cm? (Use ? = 0.05)

Transcript

00:01 For this question, there is given numbers to 4, 667 numbers we have, and we have to find the point estimate for the mean height and the standard there.

00:10 First of all, let's get the mean and the standard division for the sample.

00:14 So the sample mean, which is denoted by x bar, and the sample standard division we're going to get them.

00:22 So we're going to use the calculator, press that, and then edit.

00:26 So i'm going to just put all these values on the calculator.

00:30 4 .5 .3 and the next one is 47 .1 and enter this is 44 .2 enter and for the 6 .8 enter and we have 4 the 6 .5 enter and we have 45 .5 enter and we have 47 .6 enter and the next step again press the bottom stat in calculation there's a one variable statistics press enter and enter so we got the mean and the standard division here so the sample mean we got as, which is, so the sample mean, this is 46 points, let's say this is one.

01:07 And what about four, let me just put two decimal points, which is 14.

01:14 And the sample standard deviation, which is denoted by s, and that is equal to 1 .19.

01:23 So what we are going to do, we have to find what we have to find the standard, error here in order to get the standard error what we're going to do we're going to just find this one here so the standard error which is equal to the sample standard division divided by square of the sample size so the standard error which is 1 .19 and divided by square of there are seven numbers here because of that we just divide by square root of seven so this is 1 .19 and divided by which is the square root of seven so that would be 0 .45.

02:05 This is the standard error that we have for this question.

02:10 And the next step, so we have to calculate the margin of error.

02:14 So the margin of error, which is equal to, this is t alpha over two times the sample standard division divided by square of n.

02:21 So the alpha level was given here, which is 1 minus confidence level.

02:27 So the alpha is equal 1 minus 0 .95, which is 0 .05.

02:32 And the alpha over 2, which is equal to 0 .0 .25.

02:36 So the t alpha over 2 value, and also we need the degree of freedom, which is n minus 1.

02:41 So the degree of freedom, 7 minus 1, because there are 7 numbers, which is 6.

02:45 So that for t alpha over 2, i'm going to use the inversity function, 0 .025, and the degree of freedom is 6.

02:53 Let me just get the answer pressed second, variance and the inverse t, 0 .025, and the degree of freedom 6, which is negative 2 .45...

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