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Provided below are summary statistics for independent simple random samples from two populations. Use the pooled​...

Provided below are summary statistics for independent simple random samples from two populations. Use the pooled​ t-test and the pooled​ t-interval procedure to conduct the required hypothesis test and obtain the specified confidence interval. x overbar 1 = 11​, s 1 = 2.4​, n 1=10​, x overbar 2= 14​, s 2= 2.2​, n 2= 10 a.​ Two-tailed test, alpha =0.05 b. 95​% confidence interval (can you please show me how to input it onto calculator) ( I need help finding the P value for a the two tailed test)

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Two Tailed (non-directional) t-test 1) Calculte t_clc (t_test) 2) Find the absolute value of t calc 3) 2nd DISTR 4) Scroll down to tcdf 5) ENTER 6) Now enter: It_calcl, 1000, df) 7) ENTER 8) Output is /2 of the P-value so 9) Multiply result by 2 0

For ti-84 calculator

The following null and alternative hypotheses need to be tested This corresponds to a two-tailed test, for which a t-test for two population means with two independent samples, with unknown population standard deviations will be used (2) Rejection Region Based on the information provided, the significance level is α 0.05, and the degrees of freedom are df 18. In fact, the degrees of freedom are computed as follows, assuming that the population variances are equal Hence, it is found that the critical value for this two-tailed test is tc-2-101, for α 0.05 and df 18 The rejection region for this two-tailed test is R = {t: t> 2.101} Since it is assumed that the population variances are equal, the t-statistic is computed as follows 101) ni+n2-2 11 14 -2.914 (10-12.42+10-122 (1 10+10-2 10 10 ab Since it is observed that lt| Ξ 2.914 > tc-2. 101, it is then concluded that the null hypothesis is rejected Using the P-value approach: The p-value is p 0.0093, and since p 0.0093 < 0.05, it is concluded that the null hypothesis is rejected (5) Conclusion It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that population mean μι is different than po, at the 0.05 significance level The 95% confidence interval is 5.163 < 1- 2 <-0.837

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