Question

A test of agility involves measuring the motor skills from young adults. 20 randomly selected 20-30 year old subjects take the test and produce a mean score of 41.0 with a standard deviation of 3.7. At the 0.01 level of significance, test the claim that the true mean score for all young adults is equal to 36.0.

Ho:

H1:

test statistic:

critical value: P value:

conclusion:

Answer 1

Solution:

Here, we have to use one sample t test for the population mean.

The null and alternative hypotheses are given as below:

Null hypothesis: H0: the true mean score for all young adults is equal to 36.0.

Alternative hypothesis: Ha: the true mean score for all young adults is not equal to 36.0.

H0: µ = 36 versus Ha: µ ≠ 36

This is a two tailed test.

The test statistic formula is given as below:

t = (Xbar - µ)/[S/sqrt(n)]

From given data, we have

µ = 36

Xbar = 41

S = 3.7

n = 20

df = n – 1 = 19

α = 0.01

Critical value = - 2.8609 and 2.8609

(by using t-table or excel)

t = (Xbar - µ)/[S/sqrt(n)]

t = (41 – 36)/[3.7/sqrt(20)]

t = 6.0434

P-value = 0.0000

(by using t-table)

P-value < α = 0.01

So, we reject the null hypothesis

There is not sufficient evidence to conclude that the true mean score for all young adults is equal to 36.0.