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A researcher is concerned about the impact of students working while they are enrolled in classes, and she likes to know if students work too much and therefore are spending less time on their classes than they should be. The researcher chooses a sample of 150 students and found that the sample mean is 7.10 hours and the sample standard deviation is 5. Construct a 90% confidence interval estimate of the mean hours a week students are working. a. What is the point estimate: _____________b. Find the critical value: ________________ Round to 3 decimal places. c. What is the margin or error: __________________ Round the answer to three decimal places d. Construct the confidence interval: Express the answer as (x,x), ______________ round to one decimal place. e. State the final Conclusion.

Answer :

MrRoyal

Answer:

(a) Point estimate = 7.10

(b) The critical value is 1.960

(c) Margin of error = 0.800

(d) Confidence Interval = (6.3, 7.9)

(e) We are 90% confident that the average number of hours worked by the students is between 6.3 and 7.9

Step-by-step explanation:

Given

[tex]\bar x = 7.10[/tex] -- sample mean

[tex]\sigma=5[/tex] --- sample standard deviation

[tex]n = 150[/tex] --- samples

Solving (a): The point estimate

The sample mean can be used as the point estimate.

Hence, the point estimate is 7.10

Solving (b): The critical value

We have:

[tex]CI = 90\%[/tex] --- the confidence interval

Calculate the [tex]\alpha[/tex] level

[tex]\alpha = 1 - CI[/tex]

[tex]\alpha = 1 - 90\%[/tex]

[tex]\alpha = 1 - 0.90[/tex]

[tex]\alpha = 0.10[/tex]

Divide by 2

[tex]\frac{\alpha}{2} = 0.10/2[/tex]

[tex]\frac{\alpha}{2} = 0.05[/tex]

Subtract from 1

[tex]1 - \frac{\alpha}{2} = 1 - 0.05[/tex]

[tex]1 - \frac{\alpha}{2} = 0.95[/tex]

From the z table. the critical value for [tex]1 - \frac{\alpha}{2} = 0.95[/tex] is:

[tex]z = 1.960[/tex]

Solving (c): Margin of error

This is calculated as:

[tex]E = z * \frac{\sigma}{\sqrt n}[/tex]

[tex]E = 1.960 * \frac{5}{\sqrt {150}}[/tex]

[tex]E = 1.960 * \frac{5}{12.25}[/tex]

[tex]E = \frac{1.960 *5}{12.25}[/tex]

[tex]E = \frac{9.80}{12.25}[/tex]

[tex]E = 0.800[/tex]

Solving (d): The confidence interval

This is calculated as:

[tex]CI = (\bar x - E, \bar x + E)[/tex]

[tex]CI = (7.10 - 0.800, 7.10 + 0.800)[/tex]

[tex]CI = (6.3, 7.9)[/tex]

Solving (d): The conclusion

We are 90% confident that the average number of hours worked by the students is between 6.3 and 7.9

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