Suppose that the ""muscle gap"" in the population of all young men has a Normal distribution with standard deviation 2.5 k g / m 2 . Give a 90 % confidence interval for the mean amount of muscle young men think they should add to be attractive to women. (Enter your answers rounded to four decimal places.)

Assuming this complete 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 per square meter (/2) of fat‑free body mass. Typical young men have about 20 /2 . 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 /2 . Suppose that the “muscle gap” in the population of all young men has a Normal distribution with standard deviation 2.5 /2 . Give a 90% confidence interval for the mean amount of muscle young men think they should add to be attractive to women. (Enter your answers rounded to four decimal places.) lower limit= /2 upper limit= /2 "

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

[tex]\bar X=2.35[/tex] represent the sample mean

[tex]\mu[/tex] population mean (variable of interest)

[tex]\sigma=2.5[/tex] represent the population standard deviation

n=200 represent the sample size

Assuming the X "the gap" follows a normal distribution

[tex]X \sim N(\mu=2.35, \sigma=2.5)[/tex]

The sample mean [tex]\bar X[/tex] is distributed on this way:

[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]

The confidence interval on this case is given by:

[tex]\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex] (1)

The next step would be find the value of [tex]\z_{\alpha/2}[/tex], [tex]\alpha=1-0.90=0.1[/tex] and [tex]\alpha/2=0.05[/tex]

Using the normal standard table, excel or a calculator we see that:

[tex]z_{\alpha/2}=1.64[/tex]

Since we have all the values we can replace:

[tex]2.35 - 1.64\frac{2.5}{\sqrt{200}}=2.060[/tex]

[tex]2.35 + 1.64\frac{2.5}{\sqrt{200}}=2.640[/tex]

So on this case the 90% confidence interval would be given by (2.060;2.640)

So we are 90% confident that the true mean for the gap is between 2.06 and 2.64