Question

Construct a 95% confidence interval for the population mean of hourly wage. What sample size would you need if you wanted the confidence interval to have a width of at most $1 on either side of the sample mean (so $2 in total)?

Population mean: 5.185577655

Hourly wage:
1.878125
2.0125
2.058333
2.510417
2.752404
2.754464
3.007813
3.127404
3.188101
3.360902
3.367789
3.800481
3.822917
3.848558
3.851923
4.165865
4.357778
4.810096
5.071875
5.164063
5.5025
5.771635
5.771635
5.8675
5.985043
7.503125
8.079327
12.5025
12.65825
15.62891

Answer 1

Answer on Question #59960 – Math – Statistics and Probability

Question

Construct a 95% confidence interval for the population mean of hourly wage. What sample size would you need if you wanted the confidence interval to have a width of at most $1 on either side of the sample mean (so $2 in total)?

Population mean: 5.185577655

Hourly wage:

1.878125

2.0125

2.058333

2.510417

2.752404

2.754464

3.007813

3.127404

3.188101

3.360902

3.367789

3.800481

3.822917

3.848558

3.851923

4.165865

4.357778

4.810096

5.071875

5.164063

5.5025

5.771635

5.771635

5.8675

5.985043

7.503125

8.079327

12.5025

12.65825

15.62891

Solution

A $95\%$ confidence interval for the population mean of hourly wage is

$$C I = \left(\bar {x} - t ^ {*} \frac {s}{\sqrt {n}}; \bar {x} + t ^ {*} \frac {s}{\sqrt {n}}\right),$$

$t^{*} = 2.045$ for $95\%$ confidence level and $n - 1 = 30 - 1 = 29$ df (degrees of freedom).

See http://www.z-table.com/t-value-table.

The population mean is

$$\bar{x} = 5.139407767.$$

The sample standard deviation is

$$s = 3.283247478.$$

A 95% confidence interval for the population mean of hourly wage is

$$CI = \left(5.139407767 - 2.045 \frac{3.283247478}{\sqrt{30}}; 5.139407767 + 2.045 \frac{3.283247478}{\sqrt{30}}\right) = (3.913423010; 6.365392523)$$

The sample size would be

$$n = \left(\frac{z^* s}{E}\right)^2 = \left(\frac{1.960 \cdot 3.283247478}{1}\right)^2 = 41.4 \text{ round up to } 42.$$

Answer: (3.913423010; 6.365392523); 42.

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