Question

The average height of females in the freshman class of a certain college has historically been 162.5 centimeters with a standard deviation of 6.9 centimeters. is there reason to believe that there has been a change in the average height if a random sample of 50 females in the present freshman class has an average height of 165.2 centimeters? use a p-value in your conclusion. assume the standard deviation remains the same.

Answer 1

Explanation:

Let X be the average height of females in the freshman class.

X is N(162.5, 6.9)

Sample mean = 165.2

Sample size = 50

Since population std deviation is known and n >30 we can use z statistic

Z = (x bar-mu)/Sigma/sqrt n

i.e. Z =
`(165.2-162.5)/(6.9/√(50) )`

=2.766

Let alpha be 5%

Z critical value for 95% is 1.96

Our test statistic >1.96

p value = 0.001<0.05(our alpha)

Hence reject null hypothesis

The sample mean > the population mean

Alternate hypothesis is true.

Answer 2

To check whether there is a reason to believe that the average height has been changed, we can check one right-tailed test and establish null and alternate hypotheses.

H₀ (null hypothesis): μ = 162.5
H₁ (alternate hypothesis): μ > 162.5

Since we have two samples to compare, we can use the formula below to compute for the z-score.


`Z = ((\chi - \mu))/( \sigma / √(n))`

where Z is the z-score, Χ is the new mean of the sample, μ is the expected average value, δ is the standard deviation, and n is the sample size. Given the values we have,


`Z = ((165.2 - 162.5))/(6.9 / √(50))`

`Z \approx 2.77`

Assuming that the significance level, α, is 0.05. Now that we have a z-score of 2.77, we can use the z-table to find its p-value. Thus, we have P(Z > 2.77) = .0028. Thus since our p-value is less than α, H₀ is rejected. That means that the average height of the female freshman students has changed.