Question

Randomly selected statistics students participated in an experiment to test their ability to determine when 60 seconds has passed. Forty-six students yield a sample mean=57.43 seconds with a sample standard deviation=9.04 seconds. Use α=0.05 as a significance level to test the claim that the population mean equals to 60 seconds.

Группа выборов ответов

p-value=0.06, evidence support claim

p-value=0.6, evidence not support claim

p-value=0.06, evidence not support claim

p-value=0.06, evidence support claim

p-value=0.06, evidence not support claim

Answer 1

Solution:

Given:Randomly selected statistics students participated in an experiment to test their ability to determine when 60 seconds has passed.

Claim: the population mean equals to 60 seconds.

Sample size =n = 46

Sample mean = seconds

Sample standard deviation= s = 9.04 seconds

Significance level = α=0.05

Thus we need to use following steps;

State H0 and H1:

( claim) Vs  

this is two tailed test, since claim is non-directional.

Use following steps in TI 84 plus calculator:

Press STAT and Select TESTS

Under TESTS, select TTest

Select Stats

Under Stats Enter numbers:

Click on Calculate and press Enter.

Thus

t test statistic value = -1.928

p-value =0.06

We can use formula as follows:

t test statistic :

To find p-value ,use following Excel command:

=T.DIST.2T(x , df )

where df = n-1 = 46-1=45 and x = absolute t = 1.928

=T.DIST.2T(1.928,45)

=0.06018

=0.06

thus p-value = 0.06

Decision Rule:
Reject null hypothesis H0, if p-value < 0.05 level of significance, otherwise we fail to reject H0

Since p-value = 0.06 > 0.05 level of significance,  we fail to reject H0.

Since claim is null hypothesis and we fail to reject it, thus correct answer is:

p-value=0.06, evidence support claim