Question

In: Statistics and Probability

Required information [The following information applies to the questions displayed below.] A recent national survey found...

Required information

[The following information applies to the questions displayed below.]

A recent national survey found that high school students watched an average (mean) of 6.7 DVDs per month with a population standard deviation of 0.80 hour. The distribution of DVDs watched per month follows the normal distribution. A random sample of 40 college students revealed that the mean number of DVDs watched last month was 6.20. At the 0.05 significance level, can we conclude that college students watch fewer DVDs a month than high school students?

a. State the null hypothesis and the alternate hypothesis.

Multiple Choice

  • H0: μ ≤ 6.7 ; H1: μ > 6.7

  • H0: μ = 6.7 ; H1: μ ≠ 6.7

  • H0: μ > 6.7 ; H1: μ = 6.7

  • H0: μ ≥ 6.7 ; H1: μ < 6.7

b. State the decision rule.

Multiple Choice

  • Reject H0 if z < -1.645

  • Reject H1 if z > -1.645

  • Reject H0 if z > -1.645

  • Reject H1 if z < -1.645

c.

Compute the value of the test statistic. (Negative amount should be indicated by a minus sign. Round your answer to 2 decimal places.)

  Value of the test statistic   
d. What is your decision regarding H0?

Multiple Choice

  • Reject H0

  • Cannot reject H0

Solutions

Expert Solution

a) H0: μ ≥ 6.7 ; H1: μ < 6.7

b) Reject H0 if z < -1.645

c) population std dev ,    σ =    0.8000                  
Sample Size ,   n =    40                  
Sample Mean,    x̅ =   6.2000                  
                          
'   '   '                  
                          
Standard Error , SE = σ/√n =   0.8000   / √    40   =   0.1265      
Z-test statistic= (x̅ - µ )/SE = (   6.200   -   6.7   ) /    0.1265   =   -3.95

d)

Reject Ho


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