Grant has a sample of 36 rats in his study. If he conducts a single-sample t test, what is the value of the degrees of freedom? 37

6

36

35

35

Step-by-step explanation:

Given

Sample of rat

Population = 36

Required

Find the degree of freedom.

Given that the sample is a single-sample t test;

A single-sample t test has just one parameter has its estimate;

And its degree of freedom is calculated as thus;

Where df represents the degree of freedom and n represents the sample size;

By substituting n = 36;

Hence, the degree of freedom of the sample of 36 rats for a single-sample t test is 35

irst, find the z score of a score of 80. this is given by the formula z = (x-u)/o, where x is the score, u is the mean, and o is the standard deviation. this gives

z = (80-78)/12 = 1/6 = .1667

now, find what percentile a z score of 1/6 correlates to in the normal distribution.

a z score of 1/6 = .1667 is equivalent to the 56.36th percentile of the distribution. so the probability a score is over this is 1-.5636 = .4364

the mean of a sample mean is equal to the original mean. however, the standard deviation of a sample mean is equal to the original standard deviation divided by the square root of n (sample size). so the new standard deviation is 12/sqrt(16) = 3.

the new z score is 80-78/3 = 2/3 = .667. this is equivalent to the 74.86th percentile. so the probability that the mean will be greater than this score is 1-.7486 = .2514.

answer: no

step-by-step explanation: