Mathematics, 08.10.2020 01:01 Cartucho1978

A personnel manager has found that historically the scores on aptitude test given to applicants for entry-level positions follow a normal distribution with a standard deviation of 32.4 points. A random sample of nine test scores from the current group of applicants had a mean score of 187.9 points. a) Find an 80% confidence interval for the population mean score of the current group of applicants. b) Based on these sample results, a statistician found for the population mean a confidence interval extending from 165.8 to 210.0 points. Find the confidence level of this interval.

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Answer from: adriannacomrosenbark

Using the z-distribution, it is found that

a) The 80% confidence interval for the population mean score of the current group of applicants is (174.1, 201.7).

b) The confidence level is of approximately 96%.

Item a:

We are given the standard deviation for the population, which is why the z-distribution is used to solve this question.

The information given is:

Sample mean of $\overline{x} = 187.9$ .Population standard deviation of $\sigma = 32.4$ .Sample size of n = 9

The confidence interval is:

$\overline{x} \pm z\frac{\sigma}{\sqrt{n}}$

The critical value, using a z-distribution calculator, for a two-tailed 80% confidence interval is z = 1.28.

$\overline{x} - z\frac{\sigma}{\sqrt{n}} = 187.9 - 1.28\frac{32.4}{\sqrt{9}} = 174.1$

$\overline{x} + z\frac{\sigma}{\sqrt{n}} = 187.9 + 1.28\frac{32.4}{\sqrt{9}} = 201.7$

The 80% confidence interval for the population mean score of the current group of applicants is (174.1, 201.7).

Item b:

The margin of error is given by:

$M = z\frac{\sigma}{\sqrt{n}}$

Hence:

$\frac{210 - 165.8}{2} = z\frac{32.4}{\sqrt{9}}$

10.8z = 22.1

$z = \frac{22.1}{10.8}$

z = 2.05

Using a calculator, it is found that the confidence level is of approximately 96%.

A similar problem is given at link

Answer from: nawafcanada

A) An 80% confidence interval for the population mean score of the current group of applicants is (172.8124,202.9876)

B)The confidence level of this interval is 96%

Step-by-step explanation:

Mean = $\mu = 187.9$

Standard deviation = s= 32.4

n = 9

A) Find an 80% confidence interval for the population mean score of the current group of applicants.

Degree of freedom = n-1=9-1=8

Significance level = $\alpha = 0.2$

So, $t_(\frac{\alpha}{2},df)=t_(\frac{0.2}{2},8)=t_(0.1,8)=1.397$

Formula of confidence interval :

CI : $\bar{x}-Z_{(\frac{\alpha}{2})} \times \frac{s}{\sqrt{n}} , \bar{x}+Z_{(\frac{\alpha}{2})}\times \frac{s}{\sqrt{n}}$

So, CI: $187.9-1.397 \times \frac{32.4}{\sqrt{9}} , 187.9+1.397 \times \frac{32.4}{\sqrt{9}}$

CI:(172.8124,202.9876)

So, an 80% confidence interval for the population mean score of the current group of applicants is (172.8124,202.9876)

B)Based on these sample results, a statistician found for the population mean a confidence interval extending from 165.8 to 210.0 points. Find the confidence level of this interval.

Standard error of sample mean = se = $\frac{s}{\sqrt{n}}= \frac{32.4}{\sqrt{9}} = 10.8$

Margin of error = 210 - 165.8 = 44.2

Critical test statistic = $\frac{\text{margin of error}}{2 \times \text{standard error}} = \frac{44.2}{2 \times 10.8} = 2.0463$

Using excel

Confidence interval = $2 \times [1 - NORMSDIST(2.0463)] = 2 \times (1 - 0.98) = 2 \times 0.02 = 0.04$

Confidence level= 1 - 0.04 = 0.96

Confidence interval is 96%.

Hence the confidence level of this interval is 96%

Answer from: Quest

c. 4 and 1/2 pounds is the answer! : )

Answer from: Quest

answer: add the !

step-by-step explanation: then we will answer

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