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An association of Christmas tree growers in Indiana sponsored a sample survey of 500 randomly selected Indiana households to help improve the marketing of Christmas trees. One question the researchers asked was, "Did you have a Christmas tree this year?" Respondents who had a tree during the holiday season were asked whether the tree was natural or artificial. Respondents were also asked if they lived in an urban area or in a rural area. The tree growers want to know if there is a difference in preference for natural trees versus artificial trees between urban and rural households. Among the 160 who lived in rural areas, 64 had a natural tree. Among the 261 who lived in an urban area, 89 had a natural tree. Construct and interpret a 95% confidence interval for the difference in the proportion of rural and urban Indiana residents who had a natural Christmas tree this year.

Respuesta :

Answer:

[tex](0.4-0.341) - 1.96 \sqrt{\frac{0.4(1-0.4)}{160} +\frac{0.341(1-0.341)}{261}}=-0.0362[/tex]  

[tex](0.4-0.341) + 1.96 \sqrt{\frac{0.4(1-0.4)}{160} +\frac{0.341(1-0.341)}{261}}=0.154[/tex]  

And the 95% confidence interval would be given (-0.0362;0.154).  

We are confident at 95% that the difference between the two proportions is between [tex]-0.0362 \leq p_A -p_B \leq 0.154[/tex]

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

[tex]p_A[/tex] represent the real population proportion for rural

[tex]\hat p_A =\frac{64}{160}=0.4[/tex] represent the estimated proportion for rural  with natural tree

[tex]n_A=160[/tex] is the sample size required for rural with natural tree

[tex]p_B[/tex] represent the real population proportion for urban with natural tree

[tex]\hat p_B =\frac{89}{261}=0.341[/tex] represent the estimated proportion for urban with natural tree

[tex]n_B=261[/tex] is the sample size required for Brand B

[tex]z[/tex] represent the critical value for the margin of error  

Solution to the problem

The population proportion have the following distribution  

[tex]p \sim N(p,\sqrt{\frac{p(1-p)}{n}})[/tex]  

The confidence interval for the difference of two proportions would be given by this formula  

[tex](\hat p_A -\hat p_B) \pm z_{\alpha/2} \sqrt{\frac{\hat p_A(1-\hat p_A)}{n_A} +\frac{\hat p_B (1-\hat p_B)}{n_B}}[/tex]  

For the 95% confidence interval the value of [tex]\alpha=1-0.95=0.05[/tex] and [tex]\alpha/2=0.025[/tex], with that value we can find the quantile required for the interval in the normal standard distribution.  

[tex]z_{\alpha/2}=1.96[/tex]  

And replacing into the confidence interval formula we got:  

[tex](0.4-0.341) - 1.96 \sqrt{\frac{0.4(1-0.4)}{160} +\frac{0.341(1-0.341)}{261}}=-0.0362[/tex]  

[tex](0.4-0.341) + 1.96 \sqrt{\frac{0.4(1-0.4)}{160} +\frac{0.341(1-0.341)}{261}}=0.154[/tex]  

And the 95% confidence interval would be given (-0.0362;0.154).  

We are confident at 95% that the difference between the two proportions is between [tex]-0.0362 \leq p_A -p_B \leq 0.154[/tex]

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