15, n2 Suppose n1 respectively, and the standard deviations for samples 1 and 2 are 8.27 and 8.38, respectively. Test HO: µ1 < µ2 vs. Ha: µ1 > µ2 using a = 0.05. 12, the means for samples 1 and 2 are 70.5 and 78.5, Be sure to a. address all necessary assumptions for running a t-test; The null Hypothesis: H0: µ1 < µ2 (population 1 and 2 have equal means) The alternative Hypothesis: Ha: µ1 > µ2 (population 1 has a greater mean than population 2) A right-tailed test needs to be performed. The samples are independent of each other and o, and 0z although unknown are equal. The sample sizes are small; however, the populations are normally distributed. The assumptions for running a t-test are satisfied. b. Regardless of whether the assumptions are met, run a t-test. calculate the test statistic; Using the t-distribution for critical values table the curve for the t-statistic in the right area is 1.708 Using the TI-84 chose the 2-sample t-test, stats, pooled: yes (assumed the variances as equal) Test statistic -2.48311 Itol = 2.48311 C. calculate the p-value; TI-84, 2nd vars, tcdf, lower: -1099 , upper: 2.48, df 25 1- P(t + = 1- 0.9898850135 = 0.0101149865 - 2.48) = 0.01 p – value 0.01 < a = So, we reject the null hypothesis 0.05 d. state your conclusion; Since, \t| = 2.4831 > to.05,0.25 = 1.708 we can reject the null hypothesis. Therefore, we state with a 95% confidence there is enough evidence to arrive at the conclusion the population means are substantially different at the 5% significance level. e. construct and interpret a 95% confidence interval for the difference between the two population means.
15, n2 Suppose n1 respectively, and the standard deviations for samples 1 and 2 are 8.27 and 8.38, respectively. Test HO: µ1 < µ2 vs. Ha: µ1 > µ2 using a = 0.05. 12, the means for samples 1 and 2 are 70.5 and 78.5, Be sure to a. address all necessary assumptions for running a t-test; The null Hypothesis: H0: µ1 < µ2 (population 1 and 2 have equal means) The alternative Hypothesis: Ha: µ1 > µ2 (population 1 has a greater mean than population 2) A right-tailed test needs to be performed. The samples are independent of each other and o, and 0z although unknown are equal. The sample sizes are small; however, the populations are normally distributed. The assumptions for running a t-test are satisfied. b. Regardless of whether the assumptions are met, run a t-test. calculate the test statistic; Using the t-distribution for critical values table the curve for the t-statistic in the right area is 1.708 Using the TI-84 chose the 2-sample t-test, stats, pooled: yes (assumed the variances as equal) Test statistic -2.48311 Itol = 2.48311 C. calculate the p-value; TI-84, 2nd vars, tcdf, lower: -1099 , upper: 2.48, df 25 1- P(t + = 1- 0.9898850135 = 0.0101149865 - 2.48) = 0.01 p – value 0.01 < a = So, we reject the null hypothesis 0.05 d. state your conclusion; Since, \t| = 2.4831 > to.05,0.25 = 1.708 we can reject the null hypothesis. Therefore, we state with a 95% confidence there is enough evidence to arrive at the conclusion the population means are substantially different at the 5% significance level. e. construct and interpret a 95% confidence interval for the difference between the two population means.
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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I need help with the last part of the problem for calculating the confidence intervals. I do not even know if my calculations are correct, but I included them in case the information was needed. Thank you!
Transcribed Image Text:15, n2
Suppose n1
respectively, and the standard deviations for samples 1 and 2 are 8.27 and 8.38,
respectively. Test HO: µ1 < µ2 vs. Ha: µ1 > µ2 using a = 0.05.
12, the means for samples 1 and 2 are 70.5 and 78.5,
Be sure to
a. address all necessary assumptions for running a t-test;
The null Hypothesis: H0: µ1 < µ2 (population 1 and 2 have equal means)
The alternative Hypothesis: Ha: µ1 > µ2 (population 1 has a greater mean
than population 2)
A right-tailed test needs to be performed.
The samples are independent of each other and o, and 0z although unknown
are equal.
The sample sizes are small; however, the populations are normally
distributed.
The assumptions for running a t-test are satisfied.
b. Regardless of whether the assumptions are met, run a t-test. calculate the
test statistic;
Using the t-distribution for critical values table the curve for the t-statistic in
the right area is 1.708
Using the TI-84 chose the 2-sample t-test, stats, pooled: yes (assumed the
variances as equal)
Test statistic
-2.48311
Itol = 2.48311
C. calculate the p-value;
TI-84, 2nd vars, tcdf, lower: -1099 , upper: 2.48, df 25
1- P(t +
= 1- 0.9898850135
= 0.0101149865
- 2.48)
= 0.01
p – value 0.01 < a =
So, we reject the null hypothesis
0.05
Transcribed Image Text:d. state your conclusion;
Since, \t| = 2.4831 > to.05,0.25 = 1.708 we can reject the null hypothesis.
Therefore, we state with a 95% confidence there is enough evidence to arrive
at the conclusion the population means are substantially different at the 5%
significance level.
e. construct and interpret a 95% confidence interval for the difference
between the two population means.
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