Question

A simple random sample of 16 supermodels' heights (in centimeters) was taken.

178, 177, 176, 174, 175, 178, 175, 178, 178, 177, 190, 176, 180, 178, 180, 176

a) use a .05 significance level to test the claim that supermodels have a mean height greater than the population of women, whose mean is 162 cm.

b) Given that there are only 16 values in the sample, are your conditions from part a valid? Why or why not?

c) Construct a 95% confidence interval estimate of supermodels' mean height.

d) Compare your results from part a and c. Do they seem to agree with each other? Why do you think this is?

Answer 1

a)

Sample size = n = 16

Sample mean = = 177.875

Standard deviation = s = 3.6492

Claim: Supermodels have a mean height greater than the population of women, whose mean is 162 cm.

The null and alternative hypothesis is

Level of significance = 0.05

Here population standard deviation is unknown so we have to use t-test statistic.
Test statistic is

Degrees of freedom = n - 1 = 16 - 1 = 15

Critical value = 1.753    ( Using t table)

Test statistic > critical vaue we reject null hypothesis.

Conclusion:

Supermodels have a mean height greater than the population of women, whose mean is 162 cm.

b)

The data is a random sample.

The data has normality.

The population standard deviation is unknown.

c)

Sample size = n = 16

Sample mean = = 177.875

Standard deviation = s = 3.6492

We have to construct 95% confidence interval.

Formula is

Here E is a margin of error.

Degrees of freedom = n - 1 = 16 - 1 = 15

Level of significance = 0.05

tc = 2.131   ( Using t table)

So confidence interval is ( 177.875 - 1.9445 , 177.875 + 1.9445) = > ( 175.9305 , 179.8195)

d)

The confidence interval  (175.9305, 179.8195) which is greater than 162. So the result is the same for both parts. it seems to agree each other.