Answer to Question #214309 in Statistics and Probability for rub

Question

Before an increase in excise duty on tea , 400 people out of sample of 500 were found to be tea drinkers. After an increase in duty, 400 people were tea drinkers in a sample of 600 people. Using standard error of proportion , state whether there is significant decrease in consumption of tea at 5% level of significance. Take value of Z at 5% level of significance as 1.645.

Answer 1

Before increasing excise duty-

 $n_1=100, x_1=800$

After excise duty-

$n_2=1200,x_2=800$

Sample proportion of taking coffee before increasing excise duty-

$=p_1=\dfrac{x_1}{n_1}=\dfrac{800}{1000}=0.8$

Sample proportion of taking coffee after increasing excise duty-

$=p_2=\dfrac{x_2}{n_2}=\dfrac{800}{1200}=0.67$

Let 

Null hypothesis:$H_o:p_1=p_2=p$ i.e. there is no significant difference in the consumption of coeefe before and after the increase in excise duty.

Alternate Hypothesis:$H_1=P_1>P_2$ i.e. there is significant difference in the consumption of coeefe before and after the increase in excise duty.

Test statistics is-

$Z=\dfrac{p_1-p_2}{\sqrt{\hat{p}(1-\hat{p})(\dfrac{1}{n_1}+\dfrac{1}{n_2})}}$

Where,$\hat{p}=\dfrac{n_1p_1+n_2p_2}{n_1+n_2}=\dfrac{x_1+x_2}{n_1+n_2}=\dfrac{800+800}{1000+1200}=\dfrac{1600}{2200}=0.73$

so , $Z=\dfrac{0.8-0.67}{\sqrt{0.73\times 0.23(\dfrac{1}{1000}+\dfrac{1}{1200}})}=\dfrac{0.13}{\sqrt{0.0031}}=7.22$

Tabulated value of z at 5% level of significance for right tailed test is $1.645 i.e. Z_{0.05}=1.645$

Conclusion: Since Calculated value of Z is greater than the tabulated value of Z, It is significant and $H_o$ is rejected i.e.There has been a significant decrease in the consumption of coeffe.