Question

Question 3 After feeding a special diet to 64 mice, John measures their weights in grams and obtains the values of (sample mean) and s (sample standard deviation). He also states that a 95% confidence interval for μ (population mean) is (34.02, 35.98). (a) (2 points) Based on this information alone, determine the values of t and s

Please have fully worked out detailed explanation.

Thank you!

Answer 1

(a)

Sample size, n =64

Significance level, $\alpha$ =1 - 95% =0.05

At 0.05 significance level and n - 1= 63 degrees of freedom(df), for two-tailed case, the critical value of t is:

t =$\pm$​​​​​​1.998

Population standard deviation is unknown. So, we use t-score.

Given 95% confidence interval for the population mean, $\mu$ is: (34.02, 35.98).

Sample mean, $\bar{x}$ =(lower bound+upper bound)/2 =(34.02+35.98)/2 =35

Margin of error, MoE =(upper bound - lower bound)/2 =(35. 98 - 34.02)/2 =0.98

We know that MoE =t.s/$\sqrt{n}$

So, 0.98 =$1.998.s/\sqrt{64}$$\implies$ s =0.98(8)/1.998=3.924

Therefore, $\bar{x}$ =35 and s =3.924

(b)

Significance level, $\alpha$ =1 - 90% =0.10

At df =63 and 0.10 significance level, for two-tailed case, the critical value of t is: t=$\pm$1.669

90% confidence interval for the population mean, $\mu$ is:

$\mu =\bar{x}\ \pm t.s/\sqrt{n}$ =$\mu =35\ \pm [1.669(3.924)/\sqrt{64}]$ =(34.18, 35.82)

Comparing two confidence intervals:

90% confidence interval is (34.18, 35.82)

95% confidence interval is (34.02, 35.98)

So, when confidence level is increased from 90% to 95% the width of the interval(difference between bounds) increased, i.e., the interval became wider. Hence, 95% confidence interval is more accurate than that of 90% confidence interval as it has more chance of including the population mean value.