Question

In a random sample of 3,000 non-smckers, 270 suffer from heart disease and, in sample of 800 smokers, 88 suffer from heart disease. At the 0.01 level of significance, test whether the prevalence of heart disease is significantly higher in smokers as compared to non-smokers? [8 markah/marks] (b) Seorang penyelidik ingin membandingkan gaji purata penganalisis kewangan wanita dan lelaki. Seramai 25 orang penganalisis kewangan lelaki dan 16 penganalisis kewangan wanita yang dipilih secara rawak. Gaji purata dan sisihan piawai untuk sampel diberikan di bawah: A researcher would like to compare the mean salaries of female and male financial analysts. A total of 25 male financial analysts and 16 female financial analysts are randomly selected. The salaries average and standard deviation for both samples are given below Lelaki/Male Wanita/Female X 72 Sp -9 Pada a0.01, gunakan inferens statistik (selang keyakinan atau ujian hipotesis) untuk membuat kesimpulan tentang setiap yang berikut: 3/8

Answer 1

Answer)

Ho : P1 = P2

Ha : P1 > P2

Where, P1 = smokers

And P2 = non smokers

P1 = 88/800

P2 = 270/3000

N1 = 800 and N2 = 3000

We need to find test statistics Z .

Z = (P1 - P2)/standard error

Standard error = √p*(1-p) * √{(1/n1)+(1/n2)}

And P is combined proportion:

P = (88+270)/(800+3000)

After substituting all the values,

Z = 1.72

From z table, P(Z>1.72) = 0.0427

Given significance level is 0.01

And out p-value is 0.0427

As the p-value is greater than the given significance level, we fail to reject the null hypothesis.

We do not have enough evidence to support the claim.

B)

Here we need to compare the mean salaries

Ho : u1 - u2 = 0

Ha : u1 - u2 is not equal to 0

u1 = 77.5

u2 = 72

S1 = 6 and S2 = 9

N1 = N2 = 25

As here population standard deviation is unknown, we will use t-test.

t = (u1-u2)/standard error

Standard error = √(s1^2/n1)+(s2^2/n2)}

t = 2.5423758993656

Degrees of freedom is given by

Smaller of n1-1 and n2-1

In this case degrees of freedom is 24

From t distribution

For 24 df

And 2.542, test statistics

Required p-value is = 0.017882

As the obtained P-Value (0.017882) is more than the given significance level that is 0.01

We fail to reject the null hypothesis.