The Mathematicians from Problem 4 allege that the Statisticians spend too much t

Question

The Mathematicians from Problem 4 allege that the Statisticians spend too much time squatting and not enough time bench pressing (Mathematicians don’t like leg day). The Statisticians tell the Mathematicians they can bench just as much as they squat. Suppose that the average bench of the Statisticians is 235. Suppose also that sample standard deviation of the difference between the Statisticians’ bench and squat is 20.

1. Write down the null and alternative hypotheses.

2. What is the value of the test statistic you will use to check the null hypothesis?

3. What is the p-value of the test statistic?

4. What is your statistical decision? Use = .05.

5. What is your real-world decision?

(Problem 4 below incase needed)

Suppose we wish to determine whether or not Statisticians are better squatters than Mathematicians. We don’t really need a hypothesis test to know for sure, but the Mathematicians insisted. A random sample of 50 Statisticians and 50 Mathematicians hit the gym. The average squat of the 50 Statisticians is 256 and the average squat of the 50 Mathematicians is 235. The sample standard deviation for both the Mathematicians and Statisticians was 5. Suppose that the population standard deviations are equal but unknown, so that you can use the pooled standard error for the difference.

Explanation / Answer

Given that,
mean(x)=256
standard deviation , s.d1=5
number(n1)=50
y(mean)=235
standard deviation, s.d2 =5
number(n2)=50
null, Ho: u1 = u2
alternate, H1: u1 > u2
level of significance, = 0.05
from standard normal table,right tailed t /2 =1.661
since our test is right-tailed
reject Ho, if to > 1.661
calculate pooled variance s^2= (n1-1*s1^2 + n2-1*s2^2 )/(n1+n2-2)
s^2 = (49*25 + 49*25) / (100- 2 )
s^2 = 25
we use test statistic (t) = (x-y)/sqrt(s^2(1/n1+1/n2))
to=256-235/sqrt((25( 1 /50+ 1/50 ))
to=21/1
to=21
| to | =21
critical value
the value of |t | with (n1+n2-2) i.e 98 d.f is 1.661
we got |to| = 21 & | t | = 1.661
make decision
hence value of | to | > | t | and here we reject Ho
p-value: right tail -ha : ( p > 21 ) = 0
hence value of p0.05 > 0,here we reject Ho
ANSWERS
---------------
1.
null, Ho: u1 = u2
alternate, H1: u1 > u2
2.
test statistic: 21
4.
critical value: 1.661
decision: reject Ho
3.
p-value: 0
5.
we have enough evidence to support Statisticians are better squatters than Mathematicians

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