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

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[The following information applies to the questionsdisplayed below.]
A recent national survey found that high school students watchedan average (mean) of 6.7 DVDs per month with a population standarddeviation of 0.80 hour. The distribution of DVDs watched per monthfollows the normal distribution. A random sample of 40 collegestudents revealed that the mean number of DVDs watched last monthwas 6.20. At the 0.05 significance level, can we conclude thatcollege students watch fewer DVDs a month than high schoolstudents?
a. State the null hypothesis and the alternate hypothesis.
Multiple Choice
H₀: Î¼ â‰¤ 6.7 ; H₁: Î¼ >6.7
H₀: Î¼ = 6.7 ; H₁: Î¼ â‰ 6.7
H₀: Î¼ > 6.7 ; H₁: Î¼ =6.7
H₀: Î¼ â‰¥ 6.7 ; H₁: Î¼ <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. (Negativeamount should be indicated by a minus sign. Roundyour answer to 2 decimal places.)
Â Â Value of the test statistic Â Â
d. What is your decision regarding H₀?
Multiple Choice
Reject H₀
Cannot reject H₀

Answer & Explanation Solved by verified expert

4.0 Ratings (567 Votes)

a) H0: Î¼ â‰¥ 6.7 ; H1: Î¼ < 6.7

b) Reject H0 if z < -1.645

c) population std dev , Â Â Ïƒ = Â Â 0.8000Â Â Â Â Â Â Â Â Â Â Â Â Â Â
Sample Size ,Â Â n = Â Â 40Â Â Â Â Â Â Â Â Â Â Â Â Â Â
Sample Mean, Â Â xÌ… =Â Â 6.2000Â Â Â Â Â Â Â Â Â Â Â Â Â Â
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â
'Â Â 'Â Â 'Â Â Â Â Â Â Â Â Â Â Â Â Â Â
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â
Standard Error , SE = Ïƒ/âˆšn =Â Â 0.8000Â Â / âˆš Â Â 40Â Â =Â Â 0.1265Â Â Â Â Â
Z-test statistic= (xÌ… - Âµ )/SE = (Â Â 6.200Â Â -Â Â 6.7Â Â ) / Â Â 0.1265Â Â =Â Â -3.95

Reject Ho

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