question archive Eighty-seven fifth-grade students were interviewed

Eighty-seven fifth-grade students were interviewed

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Eighty-seven fifth-grade students were interviewed. Of those interviewed, 62 preferred hot chocolate with marshmallows while the rest preferred hot chocolate without marshmallows. Can you conclude with 99% confidence that more than half of fifth-grade students prefer hot chocolate with marshmallows? Show all of your work in the dropbox, and state your conclusion here.

 

 

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Confidence interval = (0.5874, 0.8378)

Since the interval consist of values that are all above 0.5 or 50%, then we can conclude that with 99% confidence more than half of fifth-grade students prefer hot chocolate with marshmallows

The confidence interval formula is given by;

p ± Z*√(p(1-p)/n)

p = x/n = 62/87 = 0.7126

At 99% confidence, Z = 2.58

So we have;

0.7126 ± 2.58*√(0.7126(1-0.7126)/87)

= (0.5874, 0.8378)

Since the interval consist of values that are all above 0.5 or 50%, then we can conclude that with 99% confidence more than half of fifth-grade students prefer hot chocolate with marshmallows