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Confidence Intervals Quiz 1 (25 MCQs)

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1. We plan to survey college seniors to see what proportion plan to attend grad school next year. Using 98% confidence, how many students should we survey if we wish the margin of error to be within 3%?
2. A random sample of 42 textbooks has a mean price of $ 114.50 and a standard deviation of $ 12.30. Find a point estimate for the mean price of all textbooks.
3. What is the degrees of freedom equal to for the t distribution?
4. In a survey of 104 students, it was found that 79 went to the homecoming game this year. Calculate a 99% confidence interval for p.
5. Margin of Error is equal to:
6. Construct a 95% confidence interval for the population mean, $\mu$. Assume the population has a normaldistribution. A sample of 20 college students had mean annual earnings of $ 3120 with a standard deviation of$ 677.
7. Choose the appropriate confidence interval for the situation:A random sample of 75 oak trees has a mean height of 62 ft. Assume a population standard deviation of 3.8 ft.
8. Choose the appropriate confidence interval for the situation:A random sample of 30 movie theaters showed that the average price for a movie was $ 10 with a standard deviation of $ 1.20.
9. In a crash test of 15 troopers, collision repair costs are found to have a distribution that is approximately normal, with a mean of $ 1800 and a sample standard deviation of $ 950. Construct a 99% confidence interval for the repair cost.
10. A 90% confidence interval for the mean speed driven by cars on Due West Road is given by (21, 45). What was the margin of error of this study?
11. The height of women is normally distributed with an average height of 64 inches and a standard deviation of 2.5 inches. What interval contains 99.7% of all data?
12. A sample of 30 women finds the mean height to be 62.3 in and the standard deviation to be 1.5 in. Find the margin of error of women's heights with 95% confidence.
13. Compute a 90% confidence interval for the situation:A random sample of 30 movie theaters showed that the average price for a movie was $ 10 with a standard deviation of $ 1.20.
14. A random sample of 100 WHS students found that 53 of them were in possession of a pencil. What is the p-hat for this sample?
15. Thirty randomly selected students took the calculus final. If the sample mean was 92 and the standard deviation was 9.4, construct a 99 percent confidence interval for the mean score of all students from which the sample was gathered.
16. Which confidence would result in a wider interval (assuming nothing else changed), 91% or 97%?
17. A sample of 400 racing cars showed that 80 cars cost over $ 700, 000. What is the 99% confidence interval of the true proportion of cars costing over $ 700, 000?
18. Compute a 98% confidence interval:Last year, the times for Midlo soccer games were normally distributed with a standard deviation of 2.8 minutes. A random sample of 7 soccer games this year had an average length of 94 minutes.
19. A newspaper reporter asked an SRS of 100 residentsin a large city for their opinion about the mayor'sjob performance. Using the results from the sample, the C% confidence interval for the proportion of allresidents in the city who approve of the mayor's jobperformance is 0.565 to 0.695. What is the value of C?
20. Thirty-five randomly selected students took the calculus final. If the sample mean was 92 and the standard deviation was 9.4, construct a 90 percent confidence interval for the mean score of all students.
21. A study found that in a sample of 44 subjects, 9 of agree that children under 6 shouldn't be taken to nice restaurants. Calculate a 95% confidence interval for p.
22. We want to know if students work too much (and thus spend less time on their classes). Previous studies show that the standard deviation for work time is 3 hours, and a survey of 40 students has a mean of 9.8 work hours. The 95% confidence interval is _____
23. Find the value of E, the margin of error, for c = 0.90, n = 16 and s = 2.4.
24. Use the t table to find the critical value.98% confident, df = 10
25. Why do we use a t-distribution instead of a z-distribution for means?
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