Draw inferences from the data. For this part you will need to use both the male

Question

Draw inferences from the data. For this part you will need to use both the male and female data for your variable. Complete the following:

1.Use the sample data to construct a 95% confidence interval for the mean of all males.

2.Use the sample data to construct a 95% confidence interval for the mean of all females.

3.Use the sample data to construct a 95% confidence interval for the difference of means of all males and all females. Explain if there is a difference between the two means.

4.Conduct a hypothesis test with 0.05 significance to test if the mean of all males is different from the mean of all females. Show all steps of hypothesis testing in your report including interpretation of the results.

5.Discuss if the results of the hypothesis test are different from the results of the confidence interval for the difference of two means.

HT- Female

PLEASE HELP!!!!

HT -male 70.8 66.2 71.7 68.7 67.6 69.2 66.5 67.2 68.3 65.6 63.0 68.3 73.1 67.6 68.0 71.0 61.3 76.2 66.3 69.7 65.4 70.0 62.9 68.5 68.3 69.4 69.2 68.0 71.9 66.1 72.4 73.0 68.0 68.7 70.3 63.7 71.1 65.6 68.3 66.3

HT- Female

HT 64.3 66.4 62.3 62.3 59.6 63.6 59.8 63.3 67.9 61.4 66.7 64.8 63.1 66.7 66.8 64.7 65.1 61.9 64.3 63.4 60.7 63.4 62.6 60.6 63.5 58.6 60.2 67.6 63.4 64.1 62.7 61.3 58.2 63.2 60.5 65.0 61.8 68.0 67.0 57.0

Explanation / Answer

(1) This is what we know:

n=40 , xbar=68.335, and s ( standerd error) =3.02; Here standered deviation of population is unknown

95 % confidence interval for males with (40 -1) degrees of freedom so by checking t - table we got t- value equals to = +/- 2.0227

Here mean value of height for males = xbar and standard deviation = sx.sqrt(n)

95 % confidence interval =

=xbar+- t * sx/sqrt(n) = 68.335 +- 2.0227 * 3.02/sqrt(40) = (67.37, 69.30)

Q.2 This is what we know:

n=40 , ybar=63.195, and s ( standerd error) =2.741; Here standered deviation of population is unknown so we wil perform t- test

95 % confidence interval for all females with (40 -1) degrees of freedom so by checking t - table we got t- value equals to = +/- 2.0227

Here mean value of height for females= ybar and standard deviation = sy/sqrt(n)

95% confidence interval =

=ybar+- t * sy/sqrt(n) = 63.195 +- 2.0227 * 2.741/sqrt(40) = (62.31, 64.071)

Q.3 Mean difference between popuLtion means ; here population variances are not known

Mean for difference between population =xbar - ybar = 68.335 - 63.195 = 5.14

and Standerd Error for both populations sx-y = = sqrt (sx2/ n1 + sy2/ n2) = sqrt {1/40 (3.022 + 2.752) = 0.645

Here we will use t - stastics because population variances are unknown

so dF = DF = (sx2/n1 + sy2/n2)2 / { [ (sx2 / n1)2 / (n1 - 1) ] + [ (sy2 / n2)2 / (n2 - 1) ] } = 77.28 or 77

so for dF = 77, we will calculate T - value for 95 % CI which is equal to = +/- 1.99

so 95 % confidence interval = (xbar -  ybar) +- 1.99 *  sx-y

      = 5.14 +- 1.99 * 0.645

= (3.856, 6.424)

Q.4 Null Hypothesis H0 : xbar - ybar

ALternative Hypothesis H1 : xbar - ybar 0

so Here we can use the data calculated above

Standerd Error of mean difference =  sx-y = 0.645 and dF = 77

by t - test, t- value = ( xbar - ybar) / sx-y = 5.14/ 0.645 = 7.97

and by calculatig critical value of t- stat from t - table for dF = 77, it comes = 1.99

so the test is under significant level so we can reject the null hypothesis and can say that height for man and women are different.

Q.5 No, both tests results shows same thing and we come to know that there is no negative values in q.3 confidence interval that shows that there is sure difference between 2 means.

HT -male HT - Female 70.8 64.3 66.2 66.4 71.7 62.3 68.7 62.3 67.6 59.6 69.2 63.6 66.5 59.8 67.2 63.3 68.3 67.9 65.6 61.4 63 66.7 68.3 64.8 73.1 63.1 67.6 66.7 68 66.8 71 64.7 61.3 65.1 76.2 61.9 66.3 64.3 69.7 63.4 65.4 60.7 70 63.4 62.9 62.6 68.5 60.6 68.3 63.5 69.4 58.6 69.2 60.2 68 67.6 71.9 63.4 66.1 64.1 72.4 62.7 73 61.3 68 58.2 68.7 63.2 70.3 60.5 63.7 65 71.1 61.8 65.6 68 68.3 67 66.3 57 Mean 68.335 63.195 Std. Dev. 3.019556 2.741228435

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