Question

Listed below are the amounts of net worth (in millions of dollars of the ten wealthiest celebrities in a country. Construct a 95% confidence interval. What does the result tell us about the population of all celebrities? Do the data appear to be from a normally distributed population as required? 155 153 147 147 147 147 260 206 199 156 What is the confidence interval estimate of the population mean μ? S□ million su«s□ million (Round to one decimal place as needed.)

Answer 1

Answer:

95% of confidence interval are   (144.74 , 199.23)

Step-by-step explanation:

given data  155 153 147 147 147 147 260 206 199 156

given sample n =10

$mean = ∑x / n = \frac{155+ 153+ 147 +147+ 147= 147= 260+ 206 +199+ 156 }{10}$

mean = 171.7≅ 172

x               x- mean             (x-mean)^2

155           155-172= -17        289

153           153-172 = -19       361

147           147-172  = -25      625

147           147-172 = -25       625

147           147-172 = -25       625

147           147-172 = -25       625

260          260-172 = 88      7784

206         206-172 = 34       1156

199           199-172 = 27        729

156          156-172 = -16        256

                                      ∑ (x-mean)^2 = 13,075

Sample variance  S^2 = ∑ (x-mean)^2 / n-1 = 13,075 / 10-1 =1452.77

sample standard deviation S = √variance = 38.11

95% confidence interval

The degrees of freedom = n-1 = 10-1 =9

The tabulated value 't' = 2.26 at '9' degrees of freedom at 95% level of significance

χ ±  2.26 S / √n

172 ± 2.26 (38.11)/√10

The intervals are (172 - 2.26 (38.11)/√10 , 172 + 2.26 (38.11)/√10)

    ( 172 - 27.236 , 172+27.236)

  (144.74 , 199.23)

there fore the 95% of confidence interval are   (144.74 , 199.23)

Answer 2

Using the appropriate statistical relation, the confidence interval estimate for the of netwworh of the wealthiest celebrities is (133.6 ; 209.8)

Given the samples :

• X = 155 153 147 147 147 147 260 206 199 156

Using a calculator, we could obtain the sample mean and sample standard deviation :

• Mean, μ = 171.7
• Standard deviation, σ = 38.06

The confidence interval can be defined thus :

• Mean ± standard error

• Standard Error = Tcritical × σ/√n

• Tcrit ; 95% ; df = n - 1 = 10 - 1 = 9 ; Tcritical = 2.26

Standard Error = (2.26 × 38.06/√10) = 27.20

Lower confidence boundary = 171.7 - 27.20 = 133.64

Upper confidence boundary = 171.7 + 27.20 = 209.76

Therefore, the confidence interval is (133.6 ; 209.8)

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