Question

Scientists often use mice to study human disease and physiology. Parker et al. (2016) studied over 1000 newborn mice in order to better understand the genetic loci involved in many human diseases. As part of this study, the scientists snipped up to 5 mm from the end of each mouse's tail for the purpose of DNA collection. The scientists weighed each mouse immediately after snipping its tail. The provided data set lists the weights of the mice (in grams) after snipping.
Suppose the researchers wish to determine a range of plausible values for the weights of newborn mice with recently snipped tails. Use software to construct a 95% t-confidence interval for the population mean weight of newborn mice with snipped tails.
Complete the sentence to state how the researchers should correctly interpret the t-confidence interval.
The____that___of____is between the upper and lower limits of the confidence interval.
a. the true population mean weight.
b. the true sample mean weight.
c. researchers are 95% confident
Results - Descriptive Statistics
Export
n Sample Mean Standard Min Q1 Median Q3 Max
Deviation
id1200 4.250e+4 6068 26305 4.293e+4 4.425e+4 4.568e+4 47073
bwo 1200 25.39 3.030 13.40 23.50 25.20 27.40 38.70

Answer 1

Using the t-distribution, it is found that the 95% t-confidence interval for the population mean weight of newborn mice with snipped tails is (42157, 42843).

The interpretation is: The researchers are 95% confident that the true population mean weight of newborn mice with snipped tails is between the upper and lower limits of the confidence interval.

We are given the standard deviation for the sample, which is why the t-distribution is used to solve this question.

The information given is:

Sample mean of $\overline{x} = 42500$.

Sample standard deviation of $s = 6068$.

Sample size of $n = 1200$.

The confidence interval is:

$\overline{x} \pm t\frac{s}{\sqrt{n}}$

The critical value, using a t-distribution calculator, for a two-tailed 95% confidence interval, with 1200 - 1 = 1199 df, is t = 1.96.

The interval is:

$\overline{x} - t\frac{s}{\sqrt{n}} = 42500 - 1.96\frac{6068}{\sqrt{1200}} = 42157$

$\overline{x} + t\frac{s}{\sqrt{n}} = 42500 + 1.96\frac{6068}{\sqrt{1200}} = 42843$

The 95% t-confidence interval for the population mean weight of newborn mice with snipped tails is (42157, 42843).

The interpretation is that we are 95% sure that the population mean weight is in this interval, hence:

The researchers are 95% confident that the true population mean weight of newborn mice with snipped tails is between the upper and lower limits of the confidence interval.

A similar problem is given at https://brainly.com/question/25417022