Question

3. Difference in means - Variance 1 equals variance 2, problem 1 Aa Aa Furniture World, a manufacturer of office furniture, is coming out with a new computer desk, and the factory manager must develop a job design to get it produced. She has narrowed it down to two possibilities: Design A (population 1) and Design B (population 2). She will make her decision based on production time required. She has gathered samples of 25 for each design type; data can be found in this Excel® file (Note: Reload this file each time you go through this problem). You can assume production time variances are equal. Is either design quicker? What is the relevant point estimate? What is the value of the test statistic? This test statistic has: O At distribution O A normal distribution O An F distribution O A chi-squared distribution OA standard normal distribution What is the p-value for this test? Allowing for a 10% chance of a Type I error, what is your conclusion for this test? O Do not reject the null hypothesis; there is not enough evidence to conclude that one design is quicker than the other one. O Reject the null hypothesis, and conclude that one design is quicker than the other one. Do not reject the null hypothesis, and conclude that one design is quicker than the other one. O Reject the null hypothesis; there is not enough evidence to conclude that one design is quicker than the other one.

Answer 1

Solution

What is value of test statistic?

Since it is assumed that the population variances are equal, the t-statistic is computed as follows: t= X1 - X2 (n-1)s: +(n-1) (1 + n+12-2 721 n2 6.568 – 7.56 25-1)1.36862+(25–1 2.53542 65+ 25-25-2 25 25 -1.722

This test has t distribution

P-value for the test is 0.0458

Conclusion

Reject the null hypothesis and conclude that one design is quicker than other one

Confidence Interval

The 95% confidence interval −2.151<μ1​−μ2​<0.167.

Explnantion

The provided sample means are shown below: X = 6.568 X2 = 7.56 Also, the provided sample standard deviations are: 81 = 1.3686 82 = 2.5354 and the sample sizes are ni = 25 and n2 = 25. (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: Mi = H2 Ha: Hi <H2 This corresponds to a left-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 a = 0.05, and the degrees of freedom are df = 48. In fact, the degrees of freedom are computed as follows, assuming that the population variances are equal: 0.05 and df Hence, it is found that the critical value for this left-tailed test is te = -1.677, for a = 48. The rejection region for this left-tailed test is R = {t:t< -1.677}.

(3) Test Statistics Since it is assumed that the population variances are equal, the t-statistic is computed as follows: t= X1 - X (nz-1)33+(1,-1)(+ n+n2-2 12 6.568 - 7.56 (25–11.3686-7425.-1)2.5354( + ) -1.722 (4) Decision about the null hypothesis Since it is observed that t = -1.722 <te = -1.677, it is then concluded that the null hypothesis is rejected. Using the P-value approach: The p-value is p=0.0458, and since p=0.0458 < 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 Mi is less than pl2, at the 0.05 significance level. Confidence Interval The 95% confidence interval is - 2.151 <H1 - H2 <0.167.