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ulations means, samples were collected for two independent populations where the 4) In order to test...
this is not a true or false 4) In order to test the difference in populations...
this is not a true or false
4) In order to test the difference in populations means, samples were collected for two independent populations where the variances are assumed equal and the population normally distributed. The following data resulted. Find the value of the pooled variance ad 99% CI. Population 1 Population 2 - 107 x 112 s 14 n 25 - 28
Independent random samples were selected from two quantitative populations, with sample sizes, means, and variances given...
Independent random samples were selected from two quantitative populations, with sample sizes, means, and variances given below. Population 1 2 Sample Size 39 44 Sample Mean 9.3 7.3 Sample Variance 8.5 14.82 Construct a 90% confidence interval for the difference in the population means. (Use μ1 − μ2. Round your answers to two decimal places.) __________ to ____________ Construct a 99% confidence interval for the difference in the population means. (Round your answers to two decimal places.) __________ to _____________
An analyst wants to test a hypothesis on the means of two populations. She takes samples...
An analyst wants to test a hypothesis on the means of two populations. She takes samples from each population. The variances of the population are not known, but are assumed to be equal. Statistics calculated from these samples are Sample 1 Average 1,053.6 Count 34 Sample 2 Average 1,079 Count 29 The pooled variance is calculated as 365.9 Based on this data calculate the t statistic to use in the hypothesis test. Enter your answer to 2 decimal places.
3. Test the indicated claim about the means of two populations. Assume that the two samples...
3. Test the indicated claim about the means of two populations. 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. Use the P-value method. A researcher wishes to determine whether people with high blood pressure can reduce their blood pressure, measured in mm Hg, by following a particular diet. Use a significance level of 0.01 to test the claim that the treatment group is...
Two samples each of size 20 are taken from independent populations assumed to be normally distributed...
Two samples each of size 20 are taken from independent populations assumed to be normally distributed with equal variances. The first sample has a mean of 43.5 and a standard deviation of 4.1 while the second sample has a mean of 40.1 and a standard deviation of 3.2. A researcher would like to test if there is a difference between the population means at the 0.05 significance level. What can the researcher conclude? There is not sufficient evidence to reject...
Consider the following data from two independent samples with equal population variances. Construct a 99% confidence...
Consider the following data from two independent samples with equal population variances. Construct a 99% confidence interval to estimate the difference in population means. Assume the population variances are equal and that the populations are normally distributed. x overbar 1 equals= 37.1 x overbar 2 equals= 32.8 s 1 equals= 8.68 S2 equals= 9.59 N1 equals= 15 N2 equals= 16 The 99% confidence interval is ( )(. ).
The information below is based on independent random samples taken from two normally distributed populations having...
The information below is based on independent random samples taken from two normally distributed populations having equal variances. Based on the sample information, determine the 98% confidence interval estimate for the difference between the two population means. n = 12 X1 = 57 S1 = 9 n2 = 11 X2 = 54 S2 = 8 The 98% confidence interval is $(11-12) (Round to two decimal places as needed.)
16. In order to compare the means of two populations, independent random samples are selected from...
16. In order to compare the means of two populations, independent random samples are selected from each population, with the following results. n 11, You are asked to use the pooled t-test to test 0.95, 0.049, 2 16, 20.61, s20.028 Find the observed value of the test statistic (a) 60.8873 with 25 degrees of freedom (b) t 85.0113 with 25 degrees of freedom. (c) 4.5499 with 25 degrees of freedom. (c) (d) None of the above.
The information below is based on independent random samples taken from two normally distributed populations having...
The information below is based on independent random samples taken from two normally distributed populations having equal variances. Based on the sample information, determine the 95% confidence interval estimate for the difference between the two population means. n1 14 x145 n2 13 2 47 The 95% confidence interval is s (μ1-12) s Round to two decimal places as needed)
are my answers correct? Consider the following data from two independent samples with equal population variances....
are my answers
correct?
Consider the following data from two independent samples with equal population variances. Construct a 99% confidence interval to estimate the difference in population means. Assume the population variances are equal and that the populations are normally distributed x1 = 67.9 s1 = 12.8 n1 = 10 X2 74.8 s2 = 8.1 n2 = 14 Click here to see the t-distribution table, page 1 Click here to see the t-distribution table,_page 2 The 99% confidence interval is...
this is not a true or false
4) In order to test the difference in populations means, samples were collected for two independent populations where the variances are assumed equal and the population normally distributed. The following data resulted. Find the value of the pooled variance ad 99% CI. Population 1 Population 2 - 107 x 112 s 14 n 25 - 28
3. Test the indicated claim about the means of two populations. 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. Use the P-value method. A researcher wishes to determine whether people with high blood pressure can reduce their blood pressure, measured in mm Hg, by following a particular diet. Use a significance level of 0.01 to test the claim that the treatment group is...
Two samples each of size 20 are taken from independent populations assumed to be normally distributed with equal variances. The first sample has a mean of 43.5 and a standard deviation of 4.1 while the second sample has a mean of 40.1 and a standard deviation of 3.2. A researcher would like to test if there is a difference between the population means at the 0.05 significance level. What can the researcher conclude? There is not sufficient evidence to reject...
The information below is based on independent random samples taken from two normally distributed populations having equal variances. Based on the sample information, determine the 98% confidence interval estimate for the difference between the two population means. n = 12 X1 = 57 S1 = 9 n2 = 11 X2 = 54 S2 = 8 The 98% confidence interval is $(11-12) (Round to two decimal places as needed.)
16. In order to compare the means of two populations, independent random samples are selected from each population, with the following results. n 11, You are asked to use the pooled t-test to test 0.95, 0.049, 2 16, 20.61, s20.028 Find the observed value of the test statistic (a) 60.8873 with 25 degrees of freedom (b) t 85.0113 with 25 degrees of freedom. (c) 4.5499 with 25 degrees of freedom. (c) (d) None of the above.
The information below is based on independent random samples taken from two normally distributed populations having equal variances. Based on the sample information, determine the 95% confidence interval estimate for the difference between the two population means. n1 14 x145 n2 13 2 47 The 95% confidence interval is s (μ1-12) s Round to two decimal places as needed)
are my answers
correct?
Consider the following data from two independent samples with equal population variances. Construct a 99% confidence interval to estimate the difference in population means. Assume the population variances are equal and that the populations are normally distributed x1 = 67.9 s1 = 12.8 n1 = 10 X2 74.8 s2 = 8.1 n2 = 14 Click here to see the t-distribution table, page 1 Click here to see the t-distribution table,_page 2 The 99% confidence interval is...
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