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MATH225 Week 7 Assignment (30 Q/A) / MATH 225N Week 7 Assignment: Hypothesis Test for the mean-Polution Standard Deviation known (Updated, ): Chamberlain College of Nursing $13.99
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MATH225 Week 7 Assignment (30 Q/A) / MATH 225N Week 7 Assignment: Hypothesis Test for the mean-Polution Standard Deviation known (Updated, ): Chamberlain College of Nursing
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Course
MATH225 / MATH 225
Institution
MATH225 / MATH 225
MATH225 Week 7 Assignment (30 Q/A) / MATH225N Week 7 Assignment: Hypothesis Test for the mean-Polution Standard Deviation known (Updated, ): Chamberlain College of Nursing
MATH 225 Week 7 Assignment (30 Q/A) / MATH 225N Week 7 Assignment: Hypothesis Test for the mean-Polution Standard Deviation...
,
, Question
Steve listens to his favorite streaming music service when he works out. He wonders whether
the service's algorithm does a good job of finding random songs that he will like more often
than not. To test this, he listens to 50 songs chosen by the service at random and finds that he
likes 32 of them.
Use Excel to test whether Steve will like a randomly selected song more often than not, and
then draw a conclusion in the context of the problem. Use α=0.05.
Select the correct answer below:
Reject the null hypothesis. There is sufficient evidence to conclude that Steve will like a
randomly selected song more often than not.
Reject the null hypothesis. There is insufficient evidence to conclude that Steve will like a
randomly selected song more often than not.
Fail to reject the null hypothesis. There is sufficient evidence to conclude that Steve will like a
randomly selected song more often than not.
Fail to reject the null hypothesis. There is insufficient evidence to conclude that Steve will like a
randomly selected song more often than not.
Great work! That's correct.Correct answer:
Reject the null hypothesis. There is sufficient evidence to conclude that Steve will like a
randomly selected song more often than not.
Step 1: The sample proportion is pˆ=3250=0.64, the hypothesized proportion is p0=0.5,
and the sample size is n=50.
Step 2: The test statistic, rounding to two decimal places,
is z=0.64−0.50.5(1−0.5)50‾‾‾‾‾‾‾‾‾‾‾‾√≈1.98.
Step 3: Since the test is right-tailed, entering the function =1−Norm.S.Dist(1.98,1) into
Excel returns a p-value, rounding to three decimal places, of 0.024.
Step 4: Since the p-value is less than α=0.05, reject the null hypothesis. There is sufficient
evidence to conclude that Steve will like a randomly selected song more often than not.
Null and alternative hypothesis:
H0: p = 0.5
Ha: p >= 0.5
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