Question #199161

A marketing manager of a Zambian Indigenous Company took a random sample of 2000 customers who bought a newly introduced product and found that 500 customers knew about the product. After a vigorous advertisement, another sample of 1,500 of customers indicated that 700 customers knew about the new product. Test at 5% level of significance if the advertisement increased the number of customers who know about the new product


1
Expert's answer
2021-06-08T09:14:29-0400

Sample Proportion 1 p1^=5002000=0.25\hat{p_1}=\dfrac{500}{2000}=0.25

Favorable Cases 1 X1=200X_1=200

Sample Size 1 n1=400n_1=400


Sample Proportion 2 p2^=7001500=7150.4667\hat{p_2}=\dfrac{700}{1500}=\dfrac{7}{15}\approx0.4667

Favorable Cases 2 X2=700X_2=700

Sample Size 2 n2=1500n_2=1500


The value of the pooled proportion is computed as

pˉ=X1+X2n1+n2=500+7002000+15000.342857\bar{p}=\dfrac{X_1+X_2}{n_1+n_2}=\dfrac{500+700}{2000+1500}\approx0.342857



Significance Level α=0.05\alpha=0.05


The following null and alternative hypotheses for the population proportion needs to be tested:

H0:p1p2H_0:p_1\geq p_2


H1:p1<p2H_1: p_1<p_2

This corresponds to a left-tailed test, and a z-test for two population proportions will be used.

(2) Rejection Region

Based on the information provided, the significance level is α=0.05,\alpha=0.05, and the critical value for a left-tailed test is zc=1.6449.z_c=-1.6449.

The rejection region for this lrft-tailed test is R={z:z<1.6449}.R=\{z:z<-1.6449\}.


The z-statistic is computed as follows:

z=p^1p^2pˉ(1pˉ)(1/n1+1/n2)z=\dfrac{\hat{p}_1-\hat{p}_2}{\sqrt{\bar{p}(1-\bar{p})(1/n_1+1/n_2)}}

0.250.46670.342857(10.342857)(1/2000+1/1500)\approx\dfrac{0.25-0.4667}{\sqrt{0.342857(1-0.342857)(1/2000+1/1500)}}



13.363891\approx-13.363891

Since it is observed that z=13.363891<1.6449=zc,z=-13.363891<-1.6449=z_c, it is then concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population proportion p^1\hat{p}_1 is less than p^2,\hat{p}_2, at the α=0.05\alpha=0.05 significance level.


Using the P-value approach: The p-value is p=0,p=0, and since p=0<0.05=α,p=0<0.05=\alpha, it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population proportion p^1\hat{p}_1 is less than p^2,\hat{p}_2, at the α=0.05\alpha=0.05 significance level.



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