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Let the following sample of 8 observations be drawn from a normal population with unknown mean and standard deviation: 25, 11, 12, 27,26, 22, 8,5. Use Table 2. a. Calculate the sample mean and the sample standard deviation. (Round intermediate calculations to 4 decimal places. Round "Sample mean" to 3 decimal places and "Sample standard deviation" to 2 decimal places.) Sample mean 17.000Sample standard deviation 79.42 b. Construct the 95% confidence interval for the population mean (Round "t" value to 3 decimal places and final answers to 2 decimal places.) Confidence interval 83.41 to -49.418 c. Construct the 90% confidence interval for the population mean. (Round "t" value to 3 decimal places and final answers to 2 decimal places.)Confidence interval 70.21 to -36.218 d, what happens to the margin of error as the confidence level increases from 95% to 90%?as the confidence level increases, the margin of error becomes smaller

Sagot :

For the given data

(a) sample mean is 17 and sample standard deviation = 8.91 .

(b) the 95% confidence interval is (9.551 , 24.449)  .

(c) the 90% confidence interval is (11.032 , 22.968) .  .

In the question ,

the data is given as 25, 11, 12, 27,26, 22, 8,5.

So , the number of terms (n) = 8

Part(a)

the sample mean(μ)  is = (25+11+12+27+26+22+8+5)/8 = 17

the standard deviation is calculated as :

= √[(25 - 17)² + (11 - 17)² + (12 - 17)² + (27 - 17)² + (26  - 17)² + (22 - 17)² + (8 - 17)² + (5 - 17)²)]/7

Simplifying the above data further ,

we get

= 8.91

Part(b)

we know that at 95% confidence interval , the critical value is t₀.₀₂₅,₇ = 2.3646

So , the 95% confidence interval is :

= μ ± t₀.₀₂₅,₇*s/√n

= 17 ± 2.3646*8.91/√8

= 17 ± 7.449

So , the interval is ⇒ (9.551 , 24.449) .

Part(c)

we know that at 90% confidence interval , the critical value is t₀.₀₅,₇=1.8946

So , the 90% confidence interval is :

= μ ± t₀.₂₅,₇*s/√n

= 17 ± 1.8946*8.91/√8

= 17 ± 5.968

So , the interval is ⇒ (11.032 , 22.968) .

Therefore ,(a)  the sample mean and standard deviation are 17 and 8.91

(b) the 95% interval is (9.551 , 24.449) .

(c) the 90% interval is (11.032 , 22.968) .

The given question is incomplete , the complete question is

Let the following sample of 8 observations be drawn from a normal population with unknown mean and standard deviation: 25, 11, 12, 27,26, 22, 8,5.

(a) Calculate the sample mean and the sample standard deviation.

(b) Construct the 95% confidence interval for the population mean.

(c) Construct the 90% confidence interval for the population mean.

Learn more about Sample Mean here

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