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 respected, 309 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 test 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 mean of x = 1.649 and a standard deviation of a = 0.0474. Construct a 99 percent confidence interval estimate for the mean ratio of all at this university. Does the confidence interval support the claim that this ratio is the golden mean?

Explanation / Answer

Q.1 Sample proportion of readers who like the new design = 309/540 = 0.572

so, p^ = 0.572

Now we have to calculate 90% confidence interval around p^

so Z value for 90% CI => Z 90% = +-1.645

so 90% CI = p^ + - Zcriticalsqrt [ p^ (1-p^)/ n] where n is the sample size = 540

90% CI = 0.572 +- 1.645 * sqrt [ 0.572 * 0.428 /540]

= 0.572 + - 0.0350

= (0.537, 0.607)

Yes, it appear that majority of people liked the new design as lower bound is more than 0.5.

Q.2 Sample mean Xbar = 1.649

standard deviation of sample s = 0.0474

99% confidence interval estimate for the mean ratio = Xbar +- tcritical(s/n)

here tcriticalwhen CI = 99 % so alpha = 0.005 and dF = 39-1 = 38

tcritical= +/- 2.7115

99% confidence interval estimate for the mean ratio = Xbar+- tcritical(s/n)

= 1.649 +- 2.7115 * ( 0.0474/39) = (1.6284, 1.6696)

No, The confidence interval does not support the claim this ratio is the golden mean as this doesn't contain the value of golden ration which is equal to 1.618.

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