Question

1. the mean amount spent on daycare yearly by random samples of 10 families are listed: 7,213, 13,512, 6,543, 8,256, 9,106, 12,649, 10,256, 9,553, 7,698, 10,156. 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? 3. 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: Calculate sample mean for daycare data

The formula for the sample mean $\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}$, where $n = 10$ and $x_{i}$ are the individual data - points.
$\sum_{i=1}^{10}x_{i}=7213 + 13512+6543+8256+9106+12649+10256+9553+7698+10156=95942$
$\bar{x}=\frac{95942}{10}=9594.2$

Step2: For part b (using technology, assume a statistical software or calculator)

The margin of error for a confidence interval (assuming a normal distribution and large - enough sample or known population standard deviation, if not, t - distribution for small samples) depends on the confidence level and standard deviation. Since no confidence level is specified for part b in the daycare problem, we'll focus on the bear problem for a 95% confidence interval calculation. For a sample of size $n = 30$, mean $\bar{x}=620$, and standard deviation $s = 45$, the critical value for a 95% confidence interval with $n-1=29$ degrees of freedom from the t - distribution is approximately $t_{\alpha/2}=2.045$ (using t - tables or software). The margin of error $E=t_{\alpha/2}\frac{s}{\sqrt{n}}$.
$E = 2.045\times\frac{45}{\sqrt{30}}\approx2.045\times8.216\approx16.81$

Step3: Calculate 95% confidence interval for bear population

The formula for the confidence interval is $\bar{x}\pm E$.
Lower limit $=620 - 16.81=603.19$
Upper limit $=620+16.81 = 636.81$

Step4: Explain mean and margin of error for daycare (general)

The sample mean for the daycare data (9594.2) is an estimate of the average amount spent on daycare yearly by the entire population. The margin of error represents the maximum expected difference between the sample mean and the true population mean.

Step5: Explain confidence interval for bear problem

The 95% confidence interval (603.19, 636.81) for the weight of adult male grizzly bears in the park means that we are 95% confident that the true average weight of all adult male grizzly bears in the park lies within this interval. The margin of error (16.81) contributes to the width of the interval, indicating the uncertainty in our estimate of the true population mean based on the sample.

Answer:

a. The estimated mean amount spent on daycare yearly for the population (from the sample) is 9594.2.
b. For the bear problem, the margin of error for the 95% confidence interval is approximately 16.81.
c. For the daycare problem, the sample mean is an estimate of the population mean, and the margin of error shows the uncertainty. For the bear problem, the 95% confidence interval is (603.19, 636.81), which means we are 95% confident that the true average weight of adult male grizzly bears in the park is between 603.19 and 636.81 pounds.