Mathematics, 12.11.2019 01:31 chunkymonkey090

8.105 marketing and consumer behavior in the crfa 2013 canadian chef survey, 350 professional chefs were randomly selected and asked to rate the popularity of several items, including cooking methods, hot trends, and perennial favorites.29 in the hot trends section, suppose 209 chefs selected slow cooking as the most popular preparation method, 107 chefs selected gluten-free as the most popular culinary theme, and 98 selected bite-size desserts as the most popular dessert item.

a. find a 95% confidence interval for the true proportion of canadian chefs who believe slow cooking is the most popular preparation method.

b. find a 95% confidence interval for the true proportion of canadian chefs who believe gluten-free is the most popular culinary theme.

c. find a 95% confidence interval for the true proportion of canadian chefs who believe bite-size desserts are the most popular dessert item\

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Answer from: anjah

a) The 95% confidence interval for the true proportion of Canadian chefs who believe slow cooking is the most popular preparation method is (0.5457, 0.6485).

b) The 95% confidence interval for the true proportion of Canadian chefs who believe gluten-free is the most popular culinary theme us (0.2574, 0.3540).

c) The 95% confidence interval for the true proportion of Canadian chefs who believe bite-size desserts are the most popular dessert item is (0.233, 0.327).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence interval $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

a. Find a 95% confidence interval for the true proportion of Canadian chefs who believe slow cooking is the most popular preparation method.

There are 350 professionals, so n = 350 .

209 chefs selected slow cooking as the most popular preparation method, so $\pi = \frac{209}{350} = 0.5971$ .

We have that $\alpha = 0.05$ , z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$ , so z = 1.96 .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{350}} = 0.5971 - 1.96\sqrt{\frac{0.5971*0.4029}{350}} = 0.5457$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{350}} = 0.5971 + 1.96\sqrt{\frac{0.5971*0.4029}{350}} = 0.6485$

The 95% confidence interval for the true proportion of Canadian chefs who believe slow cooking is the most popular preparation method is (0.5457, 0.6485).

b. Find a 95% confidence interval for the true proportion of Canadian chefs who believe gluten-free is the most popular culinary theme.

107 chefs selected gluten-free as the most popular culinary theme, so $\pi = \frac{107}{350} = 0.3057$

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{350}} = 0.3057 - 1.96\sqrt{\frac{0.3057*0.6943}{350}} = 0.2574$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{350}} = 0.3057 + 1.96\sqrt{\frac{0.3057*0.6943}{350}} = 0.3540$

The 95% confidence interval for the true proportion of Canadian chefs who believe gluten-free is the most popular culinary theme us (0.2574, 0.3540).

c. Find a 95% confidence interval for the true proportion of Canadian chefs who believe bite-size desserts are the most popular dessert item\

98 selected bite-size desserts as the most popular dessert item, so $\pi = \frac{98}{350} = 0.28$

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{350}} = 0.28 - 1.96\sqrt{\frac{0.28*0.72}{350}} = 0.233$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{350}} = 0.28 + 1.96\sqrt{\frac{0.28*0.72}{350}} = 0.327$

The 95% confidence interval for the true proportion of Canadian chefs who believe bite-size desserts are the most popular dessert item is (0.233, 0.327).

Answer from: Quest

55/i dont know i do it for the points

step-by-step explanation:

Answer from: Quest

536

step-by-step explanation:

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