Question

A through D please

Objective 4: Construct and Interpret a Confidence Interval for a Population Mean 9.2 Estimating a Population Mean 9.2.17 O of 1 Point Question Help A simple random sample of size n is drawn from a population that is normally distributed. The sample mean, x, is found to be 114, and the sample standard deviation, s, is found to be 10. (a) Construct an 80% confidence interval about if the sample size, n, is 23. (b) Construct an 80% confidence interval about if the sample size, n, is 18. (c) Construct a 70% confidence interval about if the sample size, n, is 23 (d) Could we have computed the confidence intervals in parts (a)-(c) if the population had not been normally distributed? Click the icon to view the table of areas under the t-distribution (a) Construct an 80% confidence interval about if the sample size, n, is 23. Lower bound: :Upper bound: (Use ascending order. Round to one decimal place as needed.) ....... . . in the addit field and then click Check Answer

Answer 1

a) n = 23

df = 23 - 1 = 22

At 80% confidence level, the critical value is t^* = 1.321

The 80% confidence interval is

$\bar x \pm t^{*} * \frac{s}{\sqrt n}$

$= 114 \pm 1.321* \frac{10}{\sqrt {23}}$

$= 114 \pm 2.75$

Lower limit = 111.2

Upper limit = 116.8

b) n = 18

df = 18 - 1 = 17

At 80% confidence level, the critical value is t^* = 1.333

The 80% confidence interval is

$\bar x \pm t^{*} * \frac{s}{\sqrt n}$

$= 114 \pm 1.333* \frac{10}{\sqrt {18}}$

$= 114 \pm 3.14$

Lower limit = 110.9

Upper limit = 117.1

c) n = 23

df = 23 - 1 = 22

At 70% confidence level, the critical value is t^* = 1.061

The 70% confidence interval is

$\bar x \pm t^{*} * \frac{s}{\sqrt n}$

$= 114 \pm 1.061* \frac{10}{\sqrt {23}}$

$= 114 \pm 2.21$

Lower limit = 111.8

Upper limit = 116.2

d) No, we could not have computed the confidence interval, if the population is not normally distributed. Because the sample sizes in parts (a) - (c) are less than 30.