Statistics: Confidence Intervals
Estimating population parameters with uncertainty
Statistics: Confidence Intervals
Estimating population parameters with uncertainty
Statistics - Grade advanced
- 1
A random sample of 64 adults has a mean systolic blood pressure of 126.4 mmHg. Assume the population standard deviation is known to be 12.8 mmHg. Construct a 95% confidence interval for the population mean systolic blood pressure.
Use x-bar ± z* times sigma divided by sqrt(n).
Because the population standard deviation is known, use a z-interval. The standard error is 12.8/sqrt(64) = 1.6, and the 95% critical value is 1.96. The margin of error is 1.96(1.6) = 3.136. The 95% confidence interval is 126.4 ± 3.136, or approximately (123.26, 129.54) mmHg. - 2
A sample of 25 batteries has a mean lifetime of 18.6 hours and a sample standard deviation of 2.5 hours. Assume battery lifetimes are approximately normally distributed. Construct a 99% confidence interval for the population mean lifetime.
Use a t critical value with n - 1 degrees of freedom.
Because the population standard deviation is unknown and n = 25, use a t-interval with 24 degrees of freedom. For 99% confidence, t* is approximately 2.797. The standard error is 2.5/sqrt(25) = 0.5, so the margin of error is 2.797(0.5) = 1.3985. The 99% confidence interval is 18.6 ± 1.3985, or approximately (17.20, 20.00) hours. - 3
A survey of 1,200 voters finds that 642 support a proposed policy. Construct a 90% confidence interval for the true proportion of voters who support the policy.
Use p-hat ± z* times sqrt(p-hat(1 - p-hat)/n).
The sample proportion is 642/1200 = 0.535. For a 90% confidence interval, z* is approximately 1.645. The standard error is sqrt((0.535)(0.465)/1200) = approximately 0.0144. The margin of error is 1.645(0.0144) = approximately 0.0237. The 90% confidence interval is approximately (0.511, 0.559). - 4
A researcher reports a 95% confidence interval for the mean difference in test scores after a tutoring program as (2.4, 7.8) points, where the difference is posttest minus pretest. Interpret this interval in context.
The researcher is 95% confident that the true mean increase in test scores after the tutoring program is between 2.4 and 7.8 points. Since the entire interval is positive, the data provide evidence that the program is associated with an average score increase. - 5
A 95% confidence interval for a population mean is (48.2, 55.6). Find the sample mean and the margin of error.
The estimate is at the center of the confidence interval.
The sample mean is the midpoint of the interval: (48.2 + 55.6)/2 = 51.9. The margin of error is half the width of the interval: (55.6 - 48.2)/2 = 3.7. Therefore, the sample mean is 51.9 and the margin of error is 3.7. - 6
A 95% confidence interval for a population proportion is (0.38, 0.46). Decide whether a hypothesized population proportion of 0.50 is plausible based on this interval. Explain your reasoning.
A hypothesized population proportion of 0.50 is not plausible at the 95% confidence level because 0.50 is outside the interval (0.38, 0.46). The interval suggests that reasonable values for the population proportion are between 0.38 and 0.46. - 7
A medical study compares recovery times for two treatments. Treatment A has n = 40, mean = 12.3 days, and standard deviation = 3.1 days. Treatment B has n = 45, mean = 14.1 days, and standard deviation = 3.6 days. Construct an approximate 95% confidence interval for mean recovery time for A minus B using a two-sample t procedure with approximate critical value t* = 2.01.
For independent samples, add the two sample variance terms inside the square root.
The estimated difference is 12.3 - 14.1 = -1.8 days. The standard error is sqrt(3.1^2/40 + 3.6^2/45) = sqrt(0.24025 + 0.288) = sqrt(0.52825) = approximately 0.727. The margin of error is 2.01(0.727) = approximately 1.46. The 95% confidence interval is -1.8 ± 1.46, or approximately (-3.26, -0.34) days. This suggests Treatment A has a lower mean recovery time than Treatment B. - 8
A class collected paired data on reaction time before and after drinking a caffeinated beverage. For 18 students, the mean paired difference, defined as before minus after, is 0.082 seconds, with a standard deviation of the differences of 0.110 seconds. Construct a 95% confidence interval for the true mean paired difference. Use t* = 2.110.
Work with the list of paired differences, not with two independent samples.
This is a paired t-interval because each student is measured twice. The standard error is 0.110/sqrt(18) = approximately 0.0259. The margin of error is 2.110(0.0259) = approximately 0.0547. The 95% confidence interval is 0.082 ± 0.0547, or approximately (0.027, 0.137) seconds. Since the interval is positive, the data suggest reaction time decreased after caffeine. - 9
An environmental scientist wants the margin of error for a 95% confidence interval for the mean nitrate level in water to be at most 0.20 mg/L. Based on prior studies, the population standard deviation is estimated as 0.90 mg/L. What minimum sample size is needed?
Always round up when finding a required sample size.
Use n = (z* sigma/E)^2 with z* = 1.96, sigma = 0.90, and E = 0.20. This gives n = (1.96(0.90)/0.20)^2 = (8.82)^2 = 77.7924. Since sample size must be a whole number and the margin of error must be at most 0.20, round up to 78. The minimum sample size is 78 water samples. - 10
A pollster wants to estimate a population proportion with 99% confidence and a margin of error no larger than 0.03. No prior estimate of the proportion is available. What minimum sample size should the pollster use?
When no estimate is available, use 0.50 for the planning proportion.
With no prior estimate, use the conservative value p* = 0.50 because it maximizes p(1 - p). For 99% confidence, z* is approximately 2.576. The sample size is n = (z*)^2 p*(1 - p*)/E^2 = (2.576)^2(0.25)/(0.03)^2 = approximately 1843.27. Rounding up, the pollster should use at least 1844 people. - 11
Explain why increasing the confidence level from 90% to 99%, while keeping the same sample data and method, makes a confidence interval wider.
A higher confidence level requires a larger critical value. A larger critical value increases the margin of error, so the confidence interval becomes wider. This reflects the need to include a broader range of plausible parameter values in order to be more confident that the interval captures the true parameter. - 12
A sample of 10 observations is used to construct a t-confidence interval for a population mean. The data are strongly right-skewed with one extreme high outlier. Explain whether the t-interval is appropriate and what the researcher should consider.
For small samples, the shape of the data matters more.
The t-interval is not very appropriate because the sample size is small and the data are strongly skewed with an extreme outlier. The t procedure is fairly robust for moderate or large samples, but with n = 10 the normality condition is important. The researcher should investigate the outlier, consider whether the population distribution is approximately normal, collect more data, or use a method such as a bootstrap interval if appropriate. - 13
A regression analysis predicting final exam score from hours studied gives a slope estimate of 4.2 points per hour with standard error 1.1. Using t* = 2.06, construct a 95% confidence interval for the true slope and interpret it.
A confidence interval for a regression slope uses estimate ± t* times standard error.
The margin of error is 2.06(1.1) = 2.266. The 95% confidence interval is 4.2 ± 2.266, or approximately (1.93, 6.47) points per hour. We are 95% confident that each additional hour studied is associated with an average increase of about 1.93 to 6.47 points in final exam score, assuming the linear regression conditions are met. - 14
A study reports a 95% confidence interval for the difference in two population proportions, p1 minus p2, as (-0.07, 0.03). What conclusion should be drawn about whether the two population proportions differ?
Because the interval includes 0, the data do not provide convincing evidence at the 95% confidence level that the two population proportions differ. Plausible values include a small negative difference, no difference, and a small positive difference. - 15
A random sample of 36 machines has a mean energy use of 5.42 kWh per hour and a sample standard deviation of 0.84 kWh per hour. A 95% t-interval is constructed as (5.136, 5.704). Identify the parameter, statistic, and one correct interpretation of the confidence level.
The confidence level describes the long-run success rate of the interval method, not the probability that this one fixed interval contains the parameter.
The parameter is the true mean energy use per hour for all machines of this type. The statistic is the sample mean, 5.42 kWh per hour. A correct interpretation of the 95% confidence level is that if many random samples of 36 machines were taken and the same method were used, about 95% of the resulting intervals would contain the true population mean.