Question

Question 3.c The breaking strength of a fabric types A and B has been tested 4 times and found A 5.3, 5.6, 6.0. 6.0 В 42, 45, 3.8, 5.3 psi_A competing new fabric) Test the claim saying average strength must be the same. Use X values with Cl, use Z/t values and use p values. State the conclusion with p values. Show calculations

Answer 1

Assuming level of significance = 0.05

The provided sample means are shown below: X1 5.725 = 4.45 Also, the provided sample standard deviations are 81 0.340342964 82 0.635085296 4 and n2-4 and the sample sizes are n1

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: $\mu_1= \mu_2​$

Ha: $\mu_1\ne \mu_2​$ ​

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=6. 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​=2.447, for α=0.05 and df=6.

The rejection region for this two-tailed test is R={t:∣t∣>2.447}.

(3) Test Statistics

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

(n1-1)s+(n2-1)s11 2 5.725-4.45 410.3403429642+4-10.6350852962 (1L 1 3.539 114 2

(4) Decision about the null hypothesis

Since it is observed that ∣t∣=3.539>tc​=2.447, it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value is p=0.0122, and since p=0.0122<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.

Confidence Interval

The 95% confidence interval is 0.393<μ1​−μ2​<2.157.

Graphically

T-Test:t3.539, p 0.0122 0.40 0.35 0.30 0.25 0.20 0.15 0.10 0.05 0.00 4.0 -3.5-3.0 -2.5 -2.0 15 1.0 -0.5 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0

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