Question

many young men in north america and europe (but not in asia) tend to think they need more muscle to be attractive. one study presented 200 young american men with 100 images of men with various levels of muscle. researchers measure level of muscle in kilograms of fat - free body mass per square meter of body surface area (kg/m²). typical young men have about 20 kg/m². each subject chose two images, one that represented his own level of body muscle and one that he thought represented \what women prefer.\ the mean gap between self - image and \what women prefer\ was 2.35 kg/m². suppose that the \muscle gap\ in the population of all young men has a normal distribution with standard deviation 2.5 kg/m². if young men thought that their own level of muscle was about what women prefer, the mean \muscle gap\ in the study would be 0. we suspect (before seeing the data) that (most) young men tend to think women prefer more muscle than they themselves have. what is the value of the test statistic z? give your answer to two decimal places.

Explanation:

Step1: Recall z - statistic formula

The formula for the z - statistic in a one - sample z - test is \(z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}\), where \(\bar{x}\) is the sample mean, \(\mu\) is the population mean, \(\sigma\) is the population standard deviation, and \(n\) is the sample size.

From the problem:

• The sample mean \(\bar{x} = 2.35\) kg/m² (the mean gap between self - image and "what women prefer" from the study).
• The population mean \(\mu=0\) (our suspected mean before seeing the data).
• The population standard deviation \(\sigma = 2.5\) kg/m².
• The sample size \(n = 200\) (since there are 200 young men in the study).

Step2: Calculate the standard error

First, we calculate the standard error \(SE=\frac{\sigma}{\sqrt{n}}\). Substitute \(\sigma = 2.5\) and \(n = 200\) into the formula:
\(SE=\frac{2.5}{\sqrt{200}}\). We know that \(\sqrt{200}=\sqrt{100\times2}=10\sqrt{2}\approx14.1421\). So \(SE=\frac{2.5}{14.1421}\approx0.1768\).

Step3: Calculate the z - statistic

Now, use the z - statistic formula \(z=\frac{\bar{x}-\mu}{SE}\). Substitute \(\bar{x}=2.35\), \(\mu = 0\), and \(SE\approx0.1768\) into the formula:
\(z=\frac{2.35 - 0}{0.1768}=\frac{2.35}{0.1768}\approx13.29\) (Wait, this seems incorrect. Wait, maybe I misread the sample mean. Wait, the problem says "the mean gap between self - image and 'what women prefer' was 2.35 kg/m²". Wait, no, wait the sample size: the study had 200 young men, each subject chose two images, but the sample size for the mean calculation is \(n = 200\)? Wait, no, maybe the sample mean is 2.35, population mean \(\mu = 0\), standard deviation \(\sigma=2.5\), sample size \(n = 200\). Wait, let's recalculate the standard error: \(\sqrt{200}\approx14.142\), so \(\frac{\sigma}{\sqrt{n}}=\frac{2.5}{14.142}\approx0.1767\). Then \(z=\frac{2.35 - 0}{0.1767}\approx13.29\)? But that seems too large. Wait, maybe the sample mean is 2.35, population mean \(\mu = 0\), standard deviation \(\sigma = 2.5\), and sample size \(n=200\). Wait, perhaps I made a mistake in the sample size. Wait, the problem says "one study presented 200 young American men with 100 images...", so the sample size \(n = 200\). Wait, but let's check again. Wait, the formula for z - statistic when the population standard deviation is known is \(z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}\). So \(\bar{x}=2.35\), \(\mu = 0\), \(\sigma = 2.5\), \(n = 200\). So \(\frac{\sigma}{\sqrt{n}}=\frac{2.5}{\sqrt{200}}=\frac{2.5}{10\sqrt{2}}=\frac{0.25}{\sqrt{2}}\approx\frac{0.25}{1.4142}\approx0.1768\). Then \(z=\frac{2.35}{0.1768}\approx13.29\). But this seems very large. Wait, maybe the sample mean is 2.35, and the population mean is 0, standard deviation 2.5, sample size 200. Alternatively, maybe I misread the sample mean. Wait, the problem says "the mean gap between self - image and 'what women prefer' was 2.35 kg/m²". So the calculation is as above. But let's check with a calculator:

\(\sqrt{200}=14.1421356\)

\(2.5\div14.1421356\approx0.1767767\)

\(2.35\div0.1767767\approx13.29\)

Wait, but maybe the sample size is 200, but the mean is 2.35, standard deviation 2.5. So the z - statistic is \(z=\frac{2.35 - 0}{2.5/\sqrt{200}}=\frac{2.35\sqrt{200}}{2.5}\). \(\sqrt{200}=10\sqrt{2}\), so \(z=\frac{2.35\times10\sqrt{2}}{2.5}=\frac{23.5\sqrt{2}}{2.5}=9.4\sqrt{2}\approx9.4\times1.4142\approx13.29\).

Answer:

\(13.29\) (Wait, but this seems very large. Maybe there is a misinterpretation. Wait, maybe the sample mean is 2.35, population mean 0, standard deviation 2.5, and sample size \(n = 200\). So the calculation is correct. So the z - statistic is approximately \(13.29\). But let's check again. Wait, maybe the sample size is 200, but the mean is 2.35, standard deviation 2.5. So the z - score is \(\frac{2.35-0}{2.5/\sqrt{200}}=\frac{2.35\sqrt{200}}{2.5}\). \(\sqrt{200}\approx14.142\), so \(2.35\times14.142 = 33.2337\), then \(33.2337\div2.5 = 13.2935\approx13.29\). So the z - statistic is approximately \(13.29\).