Question

A professor thinks that the mean number of hours that students study the night before a test is 1.75. He selected a random sample of 12 students and found that the mean number of study hours was 2.44 and the standard deviation is 1.26 hours. Test the professor’s claim at α = 0.01.

a 1-6) Give the hypotheses for H₀ (a1, a2 and a3) and H1 (a4, a5 and a6)

H₀
a1) µ or p

a2) =, ≥, ≤

a3) number

H₁
a4) µ or p

  
a5) ≠, >, <

  
a6) number

b) Calculate the test statistic. t = _______ (Round your answer to 3 decimals.)

  

c 1-3) Formulate the decision rule for the p value approach.

Reject H₀ if (c1,c2,c3)
c1) t or p

c2) > or <

  
c3) number

  

d 1-2) Give the p value (d1 − round to 2 decimals) < p < (d2 - round to 2 decimals) and make a decision (d3).
d1)

  
d2)

  
d3) reject Ho or do not reject Ho

Reject HoDo not reject HoClick for List  

e1-2) Give your conclusion. At α = .___, there (is/is not) enough evidence to conclude that the mean number of hours that students study the night before a test is not 1.75.

e1)

  
e2) is or is not

  

Answer 1

n=12 ,  = 1.75

= 2.44, s = 1.26

= 0.05

null and alternative hypothesis is

Ho: = 5.16

Ha:  > 5.16

formula for test statistics is

t =1.897

test statistics: t = 1.897

decision rule for the p value approach is,

Reject Ho if ( P-Value ) < ( 0.05 )

now calculate P-Value for this one tailed test with df= n-1 = 12- 1 =11, using following excel command we get p-value as,=TDIST(1.897,11,2)

P-Value = 0.0843

Here, ( P-Value = 0.0843) > ( = 0.05 )

Hence,

do not reject Ho

Fail to reject null hypothesis At α = 0.05 , there is not enough evidence to conclude that that the mean number of hours that students study the night before a test is not 1.75