Question

1. The annual earnings​ (in dollars) of 35 randomly selected microbiologists are shown in the data table. Use the data to​ (a) find the sample​ mean, (b) find the sample standard​ deviation, and​ (c) construct a​ 98% confidence interval for the population mean.

100,095	81,395	77,851	68,184	51,784	68,158	94,436	66,446	75,556
78,621	74,185	44,588	86,744	60,998	57,664	55,388	78,844	61,904
47,480	98,368	80,618	93,023	63,344	75,122	51,049	60,873	82,029
91,267	84,094	79,834	63,872	73,989	57,207	46,919	88,912

The monthly incomes for 12 randomly selected​ people, each with a​ bachelor's degree in​ economics, are shown on the right.

4450.37 4596.92 4366.63 4455.04 4151.09 3727.28 4283.49 4527.93 4407.87 3946.46 4023.98 4221.83

(a) Find the sample mean.

(b) Find the sample standard deviation.

c) Construct a 95​%

confidence interval for the population mean

3. Use technology to construct the confidence intervals for the population variance

and the population standard deviation

Assume the sample is taken from a normally distributed population.

c=0.98

s=38

n=15

1. Find the confidence interval for the population variance is
2. Find the confidence interval for the population standard deviation.

1.75 1.86 1.55 1.62 1.78 1.91 1.39 1.53 1.45 2.01 The number of hours of reserve capacity of 10 randomly selected automotive batteries is shown

Assume the sample is taken from a normally distributed population. Construct 90%

confidence intervals for​ (a) the population variance

and​ (b) the population standard deviation.

5. A magazine includes a report on the energy costs per year for​ 32-inch liquid crystal display​ (LCD) televisions. The article states that

14 randomly selected​ 32-inch LCD televisions have a sample standard deviation of

​$3.83. Assume the sample is taken from a normally distributed population. Construct

95% confidence intervals for​ (a) the population variance

and​ (b) the population standard deviation

6. The waiting times​ (in minutes) of a random sample of

Sample of 21 people at a bank have a sample standard deviation of

3.4 minutes. Construct a confidence interval for the population variance

and the population standard deviation

Use a 95 %level of confidence. Assume the sample is from a normally distributed population.

Answer 1

n = 35

Let x be the annual earnings (in dollars) of 35 randomly selected microbiologists

	x	(x - x bar)^2
	100095	787979356.7
	78621	43520013.18
	47480	602409408.6
	91267	370291894.4
	81395	87815078.74
	74185	4669791.341
	98368	694004755.4
	84094	145684175.8
	77851	33953579.38
	44588	752735742.2
	80618	73856320.36
	79834	60995631.4
	68184	14745830.4
	86744	216677516.8
	93023	440956741.1
	63872	66455593.12
	51784	409658814.4
	60998	121573337.6
	63344	75342920.8
	73989	3861107.101
	68158	14946187.96
	57664	206210461.6
	75122	9597418.121
	57207	219544378
	94436	502296399.3
	55388	276757494.2
	51049	439951883.5
	46919	630262531.3
	66446	31114418.68
	78844	46511990.8
	60873	124345470.1
	88912	285203530.7
	75556	12474812.08
	61904	102415007.2
	82029	100099424.7
Total	2520841	8008919017

a)

Sample mean :

Sample mean = 72024.03

b)

Sample standard deviation :.

s = 15347.85 (Round to 2 decimal)

Sample standard deviation = 15347.85

c)

Confidence level = c = 0.98

alpha = 1 - c = 1 - 0.98 = 0.02

Degrees of freedom = n - 1 = 35 - 1 = 34

98% confidence interval for the population mean is

where tc is t critical value for alpha = 0.02 and df = 34

tc from excel using function:

=TINV(0.02,34)

=2.441 (Round to 3 decimal)

98% confidence interval for the population mean is (65691.44, 78356.62)