Question

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

Answer 1

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

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)=(193)/(558)=0.346` represent the proportion of people who knew the product in New York

`p_(2)=(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) \\eq p_(2)`

The statistic for this case is given by:

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

Where
`\hat p=(X_(1)+X_(2))/(n_(1)+n_(2))=(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)((1)/(558)+(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