Question

y ar old r A theme park owner wants to know he 's drevorg boys based on the word on our than the othe Note amusement paris determine whether someone can be Parta The owner where he data on 10 years and w document for det. Det h andel penders are not the same height Show your and your in content of the problem. What does this meansbout e t con the two t enom t to go on wh a Part beberding the owner found the following expected and standard deviation for 10 year old Dor o ther of parta? 10 years 10 year old girls 52.3 53.5 53.6 53.0 56.9 51.2 48.6 51.3 53.9 54.0 57.5 53.7 55.7 54.9 57.9 56.9 59.9 56.6 54.6 52.9 52.0 10 year old boys 58.1 53.0 56.2 54.2 50.9 54.4 51.4 53.5 51.0 59.7 53.4 49.2 55.7 55.1 57.7 57.7 52.6 54.7

Answer 1

(a)

Here we have two independent samples and we need to compare mean so independent sample t test should be used.

Hypotheses are:

$H_{0}:\mu_{1}=\mu_{2}$

$H_{a}:\mu_{1}\neq \mu_{2}$

Following is the output of independent sample t test:

Hypothesis Test: Independent Groups (t-test, pooled variance) 10 year old girsl, X1 10 year old boys,X2 54.361 mean 2.828 std. dev. 18 n 54.405 2.621 22 38 df 0.0434 difference (10 year old girsl, X1 10 year old boys,X2) 7.3724 pooled variance 2.7152 pooled std. dev 0.8630 standard error of difference 0 hypothesized difference 0.05 t .9601 p-value (two-tailed) F-test for equality of variance 7.997 variance: 10 year old boys,X2 6.867 variance: 10 year old girsl, X1 1.16 F 7318 p-value

The p-value is 0.9601.

Since p-value is greater than 0.05 so there is no evidence to conclude that two genders do not have same height.

(b)

Here we know the population standard deviation so we can use z test to compare population mean. Following is the test:

Here we have х-54.405,0, -2.74, п, -22 -54.361,02-2.71, n2-18 Hypotheses are: Но : 4— м, 3 H о o Test is two tailed. Level of significance: a =0.05 Test Statistics: (-,)-(4-) ((54.405-54.361)-0)/(sqrt((2.47/22)+(0.87/18)) 0.05 п, п, Critical value: =1.96 Rejection Region: If Iz >1.96, Reject H0 Decision: Since test statistics does not lie in rejection region so we fail to reject the null hypothesis P-value: P-value-0.9601 Decision: Since p-value is not less than level of significance so we fail to reject the null hypothesis =NORMSINV(1-(0.05/2)) Excel function for critical value: -2* (1-NORMSDIST(0.05)) Excel function for p-value:

The p-value is 0.9601.

Since p-value is greater than 0.05 so there is no evidence to conclude that two genders do not have same height.

Result is same as result of part b.