Question

Liars	X1= 1.52	s1=0.32	n1= 47
Truthful	X2= 1.30	s2=0.39	n2= 4

This is a table summarizing the statistics between the number of words truthful people use vs people who are lying. To analyze this data;

• choose a t procedure, justify your choice, and perform it.
• Create a 95% confidence interval for (mean1 – mean2) and interpret this interval.
• Choose an appropriate hypothesis test and perform it (give hypothesis in words, symbols, test statistic, p-value, and conclusion)
• Interpret results.

Answer 1

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: μ1​ =μ2​

Ha:μ1​ ≠ μ2​

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.

Testing for Equality of Variances

A F-test is used to test for the equality of variances. The following F-ratio is obtained:

The critical values are FL​=0.293 and FU​=14.019, and since F=0.673, then the null hypothesis of equal variances is not rejected.

(2) Rejection Region: Based on the information provided, the significance level is α=0.05, and the degrees of freedom are df=49. 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​=2.01, for α=0.05 and df=49.

The rejection region for this two-tailed test is=R={t:∣t∣>2.01}.

(3) Test Statistics

Since it is assumed that the population variances are equal, the t-statistic is computed as follows:

TEST STATISTIC = 1.301

(4) Decision about the null hypothesis: Since it is observed that ∣t∣=1.301tc​=2.01, it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value is p=0.0002, and since p=0.0002<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 μ1​ is different than μ2​, at the 0.05 significance level.

SOLUTION B] 95% confidence interval for (mean1 – mean2)

(1.52-1.3)

is 95% CONFIDENCE INTERVAL