Question

A paint manufacturer wishes to compare the drying times of two different types of paint. Independent random samples of 11 cans of Type A and 9 cans of Type B were selected and applied to a similar surface. The drying times were recorded. Type A had a mean drying time of 68.9 with a standard deviation of 0.5. Type B had a mean drying time of 70.6 with a standard deviation of 3. Do the data provide sufficient evidence at a 0.01 significance level to conclude that the mean drying time for Type A is less than the drying time for Type B? The population variances are not equal.

Answer 1

T-test for two Means – Unknown Population Standard Deviations - Unequal Variance

The following information about the samples has been provided: a. Sample Means : Xˉ1​=68.9 and Xˉ2​=70.6 b. Sample Standard deviation: s1=0.5 and s2=3 c. Sample size: n1=11 and n2=9 (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 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) The degrees of freedom Assuming that the population variances are unequal, the degrees of freedom are given by (3a) Critical Value Based on the information provided, the significance level is α=0.01, and the degree of freedom is 8.3643. Therefore the critical value for this Left-tailed test is tc​=-2.8965. This can be found by either using excel or the t distribution table. (3b) Rejection Region The rejection region for this Left-tailed test is t<-2.8965 (4)Test Statistics The t-statistic is computed as follows: (5) The p-value The p-value is the probability of obtaining sample results as extreme or more extreme than the sample results obtained, under the assumption that the null hypothesis is true. In this case, the p-value is 0.0656 (6) The Decision about the null hypothesis (a) Using the traditional method Since it is observed that t=-1.681 > tc​=-2.8965, it is then concluded that the null hypothesis is Not rejected. (b) Using p-value method Using the P-value approach: The p-value is p=0.0656, and since p=0.0656>0.01, it is concluded that the null hypothesis is Not rejected. (7) Conclusion It is concluded that the null hypothesis Ho is Not rejected. Therefore, there is Not enough evidence to claim that the population mean μ1​ is less than μ2, at the 0.01 significance level.

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