A biologist was interested in determining whether sunflower...

A biologist was interested in determining whether sunflower seedlings treated with an extract from Vinca minor roots resulted in a different average height of sunflower seedlings than the standard height of 15.7 cm at a 0.05 level of significance. The biologist treated a random sample of n = 33 seedlings with the extract and subsequently obtained the following heights: 11.5 11.8 15.7 16.1 14.1 10.5 15.2 17.0 12.8 12.4 17.5 13.5 16.5 13.5 14.4 16.7 10.9 13.0 15.1 17.1 13.3 12.4 9.1 14.3 12.9 11.1 15.0 13.3 15.8 13.5 9.3 12.2 10.3 a) What is the rejection rule for the hypothesis? b) What is the test statistic value? c) What is the p-value?

Transcript

00:01 In the question we are given a data and we have to answer three questions.

00:04 So we are given that sample size is 33 and from the data where it is being given we can apply this formula that xy equals summation of xi divided by n.

00:16 When we use this we will get the mean comes out to be 13 .66.

00:22 Similarly when we apply the...

00:23 Now for standard deviation we know that standard deviation formula is under root of submission minus x bar whole square divided by n so when we put the values we will get the value of standard deviation as 2 .54 okay now let's calculate the test statistic the test statistic the test statistic we know the formula t equals x bar minus mu divided by s divided by under root of n so when we put the values when we put the values you will get that t equals t equals 13 .66 minus 15 .7 divided by 2 .54 and this divided by under root of 33.

01:16 So when we calculate this further we will get that this comes out to be this comes out to be negative 4 .61 minus 6 .1 minus 4 .614 then then we know that n is given to us as 33 and in the formula that degree of freedom is n minus 1 so when we put when we put the values we'll get that degree of freedom is 33 minus 1 which is 30 so degree of freedom equals 32 so if we write the hypothesis null hypothesis that mu equals 15 .7 centimeter and an alternative hypothesis that mu less than 15 .7 centimeter so we can clearly write that we can clearly like that left test has failed left test has failed so this is the first conclusion now let's move forward now p value now p value let's calculate p value so we had calculated that p equals minus of 4 .614 and degree of freedom equals 32 on left pale test left tail test and we apply this we will get that p value t value equals 0 .00.

02:46 Now since t value, this comes out to be 0 .000000 is less than 0 .05, that is given alpha.

03:00 So we can write that null hypothesis is rejective.

03:04 Null hypothesis is rejective.

03:07 So this is the second conclusion that we have drawn.

03:09 Let's move forward.

03:10 Now using critical value approach...