A magazine recently changed the design of its Layout and asked readers to commen

Question

A magazine recently changed the design of its Layout and asked readers to comment on whether they liked the new design. Of 540 readers who responded 300 said they liked it. Construct a 90 percent confidence interval estimate of the proportion of all readers who like the new design. Does it appear that most people like the new design? People have claimed that the ratio of a person's overall height to navel height equals the golden ratio which is around 1.618. To tent this, a statistics professor has each of 39 students measure his or her overall height and navel height and then the ratio. The 39 values of this ratio have a mean of x = 1.649 and a standard deviation of = 0.0474. Consider a 99 percent confidence interval estimate for the mean ratio of al at this university. Does the confidence interval support the claim that this ratio is the golden mean?

Explanation / Answer

(1)
p = 309/540 = 0.572
SE = sqrt(p*(1-p)/n) = sqrt(0.572*(1-0.572)/540) = 0.0213

For 90% CI, z-value = 1.64
Margin of error = z*SE = 1.64*0.0213 = 0.0349

Confidence Interval = p +/- ME = 0.572 +/- 0.0349 = (0.5373, 0.6072)

Confidence interval indicates more than 50% likes the design.

(2)
xbar = 1.649
s = 0.0474
SE = s/sqrt(n) = 0.0474/sqrt(39) = 0.0076

For 99% CI, z-value = 2.58
ME = z*SE = 2.58 * 0.0076 = 0.0196

Confidence interval = mean +/- ME = 1.649 +/- 0.0196 = (1.6294, 1.6686)

Confidence interval does not support the claim.

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