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Assume independent random samples are available from two populations giving information about population proportions. For the first sample assume n1= 100 and x1= 45. For the second sample, assume n2= 100 and x2= 42. Use the given sample sizes and numbers to find the pooled estimate of p̄. Round your answer to the nearest thousandth.
0.479
0.305
0.435
0.392
Choose the one alternative that best completes the statement or answers the question.
Assume independent random samples are available from two populations giving information about population proportions. For the first sample assume n1= 100 and x1= 45. For the second sample, assume n2= 100 and x2= 42. Test the null hypothesis that the population proportions are equal versus the alternative hypothesis that the proportion for population 1 is greater than the proportion for population 2. This is a one-sided test and the alternative hypothesis must be stated with p1 - p2 > 0, and the test statistic calculated accordingly. Pick the correct z value and p-value. Round your answer to the nearest thousandth.
-0.428 p-value= 0.334
0.428 p-value= 0.668
0.428 p-value= 0.334
-0.428 p-value= 0.668
Construct the indicated confidence interval for the difference between the two population means. Assume that the two samples are independent simple random samples selected from normally distributed populations. Do not assume that the population standard deviations are equal.
Two types of flares are tested and their burning times are recorded. The summary statistics are given below.
Brand X n = 35 mean = 19.4 minutes standard deviation =1.4 minutes
Brand Y n = 40 mean = 15.1 minutes standard deviation = 1.3 minutes
Construct a 95% confidence interval for the differences between the mean burning time of the brand X flare and the mean burning time of the brand Y flare.
3.2 min < µX - µY < 5.4 min
3.7 min < µX - µY < 4.9 min
3.5 min < µX - µY < 5.1 min
3.9 min < µX - µY < 4.7 min
Construct a confidence interval for µd, the population mean for a population of paired differences. Assume that the population of paired differences is normally distributed. Let the sample mean calculated from a random sample of paired differences be d.
If d = 3.125, sd = 2.911, and n = 8, determine a 95 percent confidence interval for µd.
0.691 < µd < 5.559
2.264 < µd < 5.559
2.264 < µd < 3.986
0.691 < µd< 3.986
One-way Analysis of Variance is used to determine statistically whether the variance between the treatment level means is greater than the variances within levels (error variance). It is assumed that the observations are random samples drawn from normally distributed populations that have equal variances. (Refer to Business Statistics Section 11.2.)
True
False
use the given data to answer the question.
Identify the value of the test statistic.
DF SS MS F p
Factor 3 13.500 4.500 5.17 0.011
Error 16 13.925 0.870
Total 19 27.425
5.17
13.500
4.500
0.011
A manager at a bank is interested in comparing the standard deviation of the waiting times when a single waiting line is used versus when individual lines are used. He wishes to test the claim that the population standard deviation for waiting times when multiple lines are used is greater than the population standard deviation for waiting times when a single line is used. This is a test of differences in variability. Find the p-value for a test of this claim given the following sample data. If you use R, you will be able to calculate a precise p-value. If you do a table lookup, you won't be able to find the exact p-value, but will be able to bound the p-value. Retain at least 4 digits in the calculated p-value if you use R.
Sample 1: multiple waiting lines: n1 = 13, s1 = 2.1 minutes
Sample 2: single waiting line: n2 = 16, s2 = 1.1 minutes
0.05 <= p-value < 0.1
0.0205 <= p-value < 0.05
0.01 <= p-value < 0.0205
0.005 <= p-value < 0.01
Provide an appropriate response.
Fill in the missing entries in the following partially completed one-way ANOVA table.
Source df SS MS-SS/df F-statistic
Treatment 22.2
Error 26 4
Total 31
Source df SS MS-SS/df F-statistic
Treatment 5 22.2 4.44 1.11
Error 26 104.0 4
Total 31 126.2
Source df SS MS-SS/df F-statistic
Treatment 5 22.2 4.44 0.90
Error 26 104.0 4
Total 31 126.2
Source df SS MS-SS/df F-statistic
Treatment 57 22.2 0.39 316.35
Error 26 104.0 4
Total 31 126.2
Source df SS MS-SS/df F-statistic
Treatment 5 22.2 4.44 1.11
Error 26 104.0 4
Total 31 22.35
use the given data to find the equation of the regression line. Round the final values to three significant digits, if necessary. Let x be the independent variable and y the dependent variable. (Note that if x = 2, then y = 7 and so forth. yhat is the predicted value of the fitted equation.)
x 2 4 5 6
y 7 11 13 20
yhat = 0.15 + 2.8x
yhat = 3.0x
yhat = 0.15 + 3.0x
yhat = 2.8x
Find the value of the linear correlation coefficient r using the data below. ( Note that if x = 47.0 then y = 8, and so forth.)
x 47.0 46.6 27.4 33.2 40.9
y 8 10 10 5 10
0
-0.175
0.175
0.156
