Question

4.26 Measurement of suspended particles in g/m can be made for air quality monitoring Let i and pay be the average concentration of suspended particles in the city center of Melbourne and Houston, respectively. Using n = 13 observations for Melbourne and n2 = 16 observations for Houston, we test Ho: Mi = 12 versus H: Mi < 2 (a) Assuming the equal variances, find the critical region with a = 0.05. (b) Suppose that X = 72.9. Si = 25.6. Y = 81.7 and S2 = 28.3 for Melbourne and Houston, respectively. Then calculate the test statistic, and state the conclusion.

Answer 1

Solution-

a)

Based on the information provided, the significance level is a = 0.05, and the degrees of freedom are df = 27. 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 left-tailed test is tc = -1.703, for a = 0.05 and df = 27. The rejection region for this left-tailed test is R= {t:t< -1.703}.

b) where sample means are denoted by X1 and X2.

The provided sample means are shown below: X1 = 72.9 X2 = 81.7 Also, the provided sample standard deviations are: $1 = 25.6 82 = 28.3 and the sample sizes are n = 13 and n2 = 16. (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: M1 = 12 Ha: Mi < u2 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 = 27. 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 left-tailed test is tc = -1.703, for a = 0.05 and df = 27. The rejection region for this left-tailed test is R = {t:t < -1.703}.

(3) Test Statistics Since it is assumed that the population variances are equal, the t-statistic is computed as follows: X1 - X2 t = -1) s +(12-1)s ( 1 + mo) ni+n2-2 72.9 – 81.7 = -0.869 /(13-1)25.62+(16-1)28.32 ( 1 ) 13+16–2 ****(13 + a) (4) Decision about the null hypothesis Since it is observed that t = -0.869 > tc = -1.703, it is then concluded that the null hypothesis is not rejected. Using the P-value approach: The p-value is p = 0.1964, and since p=0.1964 > 0.05, it is concluded that the null hypothesis is not rejected. (5) Conclusion It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population mean Mi is less than 42, at the 0.05 significance level. Confidence Interval The 95% confidence interval is – 29.588 < Mi – M2 < 11.988.

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