Question

theme park owner wants to know if the children’s rides are favoring 10 year old girls over 10 year old boys based on height. In other words, does one group taller than the other and thus can go on more rides?

Part a: The owner gathered height data on 10 year old girls and 10 year old boys, see data below. Determine at the 5% significance level if there is evidence that the two genders are not the same height and thus cannot all go on the same rides. Show your work and give your answer in complete sentences in context of the problem.

10 year old girls	10 year old boys
52.3	58.1
53.5	53.0
53.6	56.2
53.0	54.2
56.9	50.9
51.2	54.4
48.6	51.4
51.3	53.5
53.9	51.0
54.0	59.7
57.5	53.4
53.7	49.2
55.7	55.7
54.9	55.1
57.9	57.7
56.9	57.7
59.9	52.6
56.6	54.7
54.6
52.9
52.0
56.0

Part b: After doing some research, the owner found the following information about the expected height and standard deviation for 10 year olds. Does that change the result of part a? Show your work and give your answer in complete sentences in context of the problem.

	10 year old girls	10 year old boys
Average height	54.5 inches	54.5 inches
St. deviation	2.74 inches	2.71 inches

Answer 1

Part a

Solution:

Two sample t test for population means assuming pooled variance

Null hypothesis: H₀: The mean height of 10 year old girls is same as the mean height of 10 year old boys.

Alternative hypothesis: H_a: The mean height of 10 year old girls is not same as the mean height of 10 year old boys.

H₀: µ_boy = µ_girl v/s H₀: µ_boy ≠ µ_girl (Two tailed test)

We are given level of significance as 5% or α = 0.05.

Test statistic formula for pooled variance t test is given as below:

t = (X1bar – X2bar) / sqrt[S_p²*((1/n1)+(1/n2))]

Where S_p² is pooled variance

S_p² = [(n1 – 1)*S1^2 + (n2 – 1)*S2^2]/(n1 + n2 – 2)

From given data, we are given

X1bar = 54.40455

X2bar = 54.36111

S1 = 2.620519

S2 = 2.827832

n1 = 22

n2 = 18

df = n1 + n2 – 2 = 22 + 18 – 2 = 38

α = 0.05

Critical value = -2.0244 and 2.0244

(By using t-table or excel)

S_p² = [(22 – 1)* 2.620519^2 + (18 – 1)* 2.827832^2]/(22 + 18 – 2)

S_p² = 7.3724

t = (54.40455 – 54.36111) / sqrt[7.3724*((1/22)+(1/18))]

t = 0.0434 / sqrt[7.3724*((1/22)+(1/18))]

t = 0.0503

P-value = 0.9601

(By using t-table or excel)

P-value > α = 0.05

So, we do not reject the null hypothesis

There is sufficient evidence to conclude that the mean height of 10 year old girls is same as the mean height of 10 year old boys.

Part b

Solution:

Two sample t test for population means assuming pooled variance

Null hypothesis: H₀: The mean height of 10 year old girls is same as the mean height of 10 year old boys.

Alternative hypothesis: H_a: The mean height of 10 year old girls is not same as the mean height of 10 year old boys.

H₀: µ_boy = µ_girl v/s H₀: µ_boy ≠ µ_girl (Two tailed test)

We are given level of significance as 5% or α = 0.05.

Test statistic formula for pooled variance t test is given as below:

t = (X1bar – X2bar) / sqrt[S_p²*((1/n1)+(1/n2))]

Where S_p² is pooled variance

S_p² = [(n1 – 1)*S1^2 + (n2 – 1)*S2^2]/(n1 + n2 – 2)

From given data, we are given

X1bar = 54.5

X2bar = 54.5

S1 = 2.74

S2 = 2.71

n1 = 22

n2 = 18

df = n1 + n2 – 2 = 22 + 18 – 2 = 38

α = 0.05

Critical value = -2.0244 and 2.0244

(By using t-table or excel)

t = (54.5 – 54.5) / sqrt[S_p²*((1/n1)+(1/n2))]

t = 0 / sqrt[S_p²*((1/n1)+(1/n2))]

t = 0.00

P-value = 1.00

(By using t-table or excel)

P-value > α = 0.05

So, we do not reject the null hypothesis

There is sufficient evidence to conclude that the mean height of 10 year old girls is same as the mean height of 10 year old boys.

Results for part b do not change. Results for both parts are same.