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soldi70 [24.7K]
3 years ago
5

Assume that the samples are independent and that they have been randomly selected. A marketing survey involves product recogniti

on in New York and California. Of 558 New Yorkers surveyed, 193 knew the product while 196 out of 614 Californians knew the product.
At the 0.05 significance level, test the claim that the recognition rates are the same in both states
Mathematics
1 answer:
zhuklara [117]3 years ago
4 0

Answer:

z=\frac{0.346-0.319}{\sqrt{0.332(1-0.332)(\frac{1}{558}+\frac{1}{614})}}=0.98    

Now we can calculate the p value with this probability:

p_v =2*P(Z>0.98)= 0.327    

Since the p value is higher than the significance level we have enough evidence to FAIL to reject the null hypothesis and we don't have enough evidence to conclude that the two recognition rates are different

Step-by-step explanation:

Information given

X_{1}=193 represent the number of people who knew the product in New York

X_{2}=196 represent the number of people who knew the product in California

n_{1}=558 sample 1 from New York

n_{2}=614 sample 2 from California

p_{1}=\frac{193}{558}=0.346 represent the proportion of people who knew the product in New York

p_{2}=\frac{196}{614}=0.319 represent the proportion of people who knew the product in California

\hat p represent the pooled estimate of p

z would represent the statistic

p_v represent the value for the test

\alpha=0.05 significance level

System of hypothesis

We want to verify if the recognition rates are the same in both states, the system of hypothesis are:    

Null hypothesis:p_{1} = p_{2}    

Alternative hypothesis:p_{1} \neq p_{2}    

The statistic for this case is given by:

z=\frac{p_{1}-p_{2}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{1}}+\frac{1}{n_{2}})}}   (1)  

Where \hat p=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}=\frac{193+196}{558+614}=0.332  

The statistic for this case is given by:

z=\frac{0.346-0.319}{\sqrt{0.332(1-0.332)(\frac{1}{558}+\frac{1}{614})}}=0.98    

Now we can calculate the p value with this probability:

p_v =2*P(Z>0.98)= 0.327    

Since the p value is higher than the significance level we have enough evidence to FAIL to reject the null hypothesis and we don't have enough evidence to conclude that the two recognition rates are different

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