Find the sample proportion of students

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Reference no: EM132332242

Confidence Intervals for Population Means Assignment -

1. In a random sample of n = 25 randomly selected juice drinks, the mean number of calories was x^- =100. The sample standard deviation was s = 35. Follow the steps below to construct a 95% confidence interval for the population mean number of calories in juice drinks.

(a) Calculate s/√n.

(b) Since n = 25, degrees of freedom.

(If your calculator has the invT function, which is similar to the invNorm function, you may also use your graphing calculator: Go to 2^nd (DISTR VARS) and choose 4: invT( . Type in the value of 1- α/2 , which is the area to the left of t_(α/2), and the degrees of freedom separated by commas, close the parentheses, and ENTER. )

(d) Use the formula E = t_(α/2)*s/√n to calculate the margin of error. (Round to 3 decimal places.)

(e) Once you have the margin of error, E, the endpoints confidence interval are (x^- -E, x^- +E). Find a 95% confidence interval for the population mean number of calories in juice drinks.

(f) Use the TInterval procedure on your calculator to check your answer. Are the results about the same?

(g) Interpret your confidence interval using the template on the notes page.

2. In a random sample of 50 refrigerators, the mean repair cost was $157.50, and the standard deviation was $17.25.

(a) Construct a 90% confidence interval for the population mean refrigerator repair costs. (Use the TInterval procedure on your graphing calculator.)

(b) Interpret this confidence interval.

(d) Which interval is wider?

3. The GPA's of 12 randomly selected college students are given below.

3.4 2.5 3.6 1.7 2.0 3.8 2.5 2.8 3.0 3.2 3.3 3.7

(a) Construct a 90% confidence interval for the population mean GPA. You may use the TInterval procedure.

(b) Interpret this interval.

Confidence Intervals for Proportions Assignment -

1. In a survey of n = 2500 college students, x = 1000 students reported that they worked 20 or more hours a week while attending school. Follow the steps below to construct a 95% confidence interval for the population proportion of college students who work 20 or more hours while attending school.

(a) Find the sample proportion of students who reported working 20 or more hours a week while attending school. p ^=x/n=

(b) Calculate: √((p ^(1-p ^))/n) (Round your answer to 4 decimal places.)

(d) Find the margin of error E = z_(α/2)*√((p ^(1-p ^))/n) (Round your answer to 4 decimal places.)

(e) Now find the bounds of the confidence interval: p ^±E

(f) Check your answer using the 1-PropZInterval procedure on your calculator. Do the results agree?

(g) Interpret the confidence interval using the template on the notes page.

2. In a survey of n = 1750 workers, p ^=15% reported being dissatisfied with their working conditions.

(a) How many of the 1750 workers reported being dissatisfied with their working conditions? (Round to the nearest whole number.) x = np ^=

(b) Construct a 99% confidence interval for the population proportion of workers who are dissatisfied with their working conditions. (You may use the 1-PropZInterval procedure on your calculator.)

3. You wish to estimate, with 95% confidence, the population proportion of college students who have used at least one e-book for a college course. Your estimate must be accurate within 3% of the population proportion. (m = 0.03). (Use the formulas on p. 320 of your textbook.)

(a) How many students would you need to survey if no prior estimates are available?

(b) How many students would you need to survey if a prior study showed that 40% of students have used at least one e-book for a college course?

Reference no: EM132332242

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