Consider the following hypothesis test: H0 : ? = 110 Ha : ?...

Consider the following hypothesis test: H0 : ? = 110 Ha : ? ? 110 A sample of 85 is used. Identify the p-value and state your conclusion for each of the following sample results. Use ? = 0.02. Round the value of the test statistic to three decimal places. If required enter negative values as negative numbers. a. x? = 114 and s = 10.5 t-value 3.51 p-value lower than .01 conclusion reject H0 b. x? = 104 and s = 11.5 t-value -0.57 p-value lower than .01 conclusion reject H0 c. x? = 113 and s = 12.5 t-value 2.18 p-value between .025 and .05 conclusion do not reject H0

Transcript

00:01 So consider the following hypothesis test.

00:03 So we have the 80 is equal to mu equals to 80.

00:07 And the alternative is the different sign.

00:12 So mu is different than 80.

00:14 We have a sample size of 50.

00:16 And we want a level significance equals to 0 .01.

00:22 So in letter a, we want to find what is the test or the t value, but it's the test statistic for when we are assuming that the simple means equals to 83 and the same percent deviation is equals to 10.

00:40 So in this case, we can use a t value here because it is related to the t student distribution.

00:49 But in this case, the formula is given by x more minus the value that we are testing our hypothesis divided by s divided by the square root of x.

00:59 So the value that we are testing here is this number here in the new hypothesis.

01:06 So we are testing if the population mean is equal to 80.

01:10 So this number here should be equals to the number that we are testing in our hypothesis.

01:17 So if you just plug the other information that we also have from the question, we are going to find that the t value is equal to 2 .12.

01:27 Now the p value, since this test is a two -tail or two -sided test, this is given by the probability of the module of a t -student distribution being greater than the level or the t -value that we found, which is 2 .12.

01:48 So if you open the module here, you're going to find that this is the sum of two areas in the t -student, which is given.

01:58 By the t student less than the negative of the t value plus the area in which the t student is greater than the positive value of the t value.

02:11 And now we can assume that the t, not assume, but it's a fact.

02:16 The t student is symmetric.

02:18 So these two probabilities are equal.

02:22 So we can just compute two times one of them and say that this is the t, the t's symmetric.

02:28 Value that we want.

02:29 So let's choose the first one.

02:31 And to compute this probability, we need to specify what is the degrees of freedom of this t student.

02:38 So the degrees of freedom of this t student is the sample size minus one.

02:44 So we're assuming that this t student has 49 degrees of freedom.

02:49 And assuming this, you are going to find using a t -table that this probability is equal to 0 .0195, which is 0 .039.

03:01 So in this case, we can say that this number belongs, in this case, between 0 .025 and 0 .05...

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