Question

1. the mean amount spent on daycare yearly by random samples of 52 families are listed: 13,512; 6,543; 9,250; 9,100; 12,645; 10,250; 6,543; 7,890; 10,150

a. use the values to estimate the mean amount spent on daycare yearly for the population.

b. use technology to find the associated margin of error.

c. explain in sentences what the average and margin of error mean. in other words, what is the range of plausible values for the mean amount spent on daycare by the entire population?

1. a wildlife biologist wants to estimate the average weight of adult male grizzly bears in a national park. he captures and weighs a random sample of 30 adult male grizzly bears. the sample has a mean weight of 620 pounds with a standard deviation of 45 pounds. the biologist wants to determine a 95% confidence interval for the true average weight of all adult male grizzly bears in the park.

a. using the provided sample data, calculate the 95% confidence interval for the average weight of the grizzly bear population. in other words, what is the range of values which he can expect 95% of the bear population to weigh?

b. explain what the calculated confidence interval means in the context of this problem and how the margin of error contributes to this estimate.

Explanation:

Step1: Identify the formula for confidence interval

For a 95% confidence interval when the population standard - deviation is unknown (we use the sample standard - deviation $s$ instead), the formula is $\bar{x}\pm t_{\alpha/2}\frac{s}{\sqrt{n}}$, where $\bar{x}$ is the sample mean, $t_{\alpha/2}$ is the critical value, $s$ is the sample standard deviation, and $n$ is the sample size. Here, $\bar{x} = 620$, $s = 45$, and $n=30$. The degrees of freedom $df=n - 1=30 - 1 = 29$. For a 95% confidence interval, $\alpha=1 - 0.95 = 0.05$ and $\alpha/2=0.025$. Using a $t$ - distribution table or technology, $t_{0.025,29}\approx 2.045$.

Step2: Calculate the margin of error

The margin of error $E=t_{\alpha/2}\frac{s}{\sqrt{n}}$. Substitute the values: $E = 2.045\times\frac{45}{\sqrt{30}}\approx2.045\times\frac{45}{5.477}\approx2.045\times8.216\approx16.8$.

Step3: Calculate the confidence interval

The lower limit of the confidence interval is $\bar{x}-E=620 - 16.8 = 603.2$ and the upper limit is $\bar{x}+E=620 + 16.8 = 636.8$.

Answer:

a. The 95% confidence interval for the average weight of the grizzly - bear population is $(603.2,636.8)$ pounds.
b. The calculated 95% confidence interval $(603.2,636.8)$ means that if we were to take many different samples of 30 adult male grizzly bears from the park and calculate a 95% confidence interval for each sample, about 95% of those intervals would contain the true average weight of all adult male grizzly bears in the park. The margin of error of 16.8 pounds represents the maximum amount by which the sample mean is likely to differ from the true population mean. In other words, we are 95% confident that the true average weight of adult male grizzly bears in the park is within 16.8 pounds of the sample mean of 620 pounds.