Question

An article appeared in an Australian newspaper. It described a study of academic performance and attention span and reported that the mean time to distraction for teenage boys working on an independent task was 5 minutes. Although the sample size was not given in the article, suppose that this mean was based on a random sample of 50 teenage Australian boys and that the sample standard deviation was 1.5 minutes.

Is there convincing evidence that the average attention span for teenage boys is less than 6 minutes? Test the relevant hypotheses using

α = 0.01.

(Use a statistical computer package to calculate the P-value. Round your test statistic to two decimal places and your P-value to three decimal places.)

t=

P-value=

State the conclusion in the problem context.

Do not reject H₀. We have convincing evidence that the mean time to distraction for Australian teenage boys is less than 6 minutes.Reject H₀. We do not have convincing evidence that the mean time to distraction for Australian teenage boys is less than 6 minutes.    Reject H₀. We have convincing evidence that the mean time to distraction for Australian teenage boys is less than 6 minutes.Do not reject H₀. We do not have convincing evidence that the mean time to distraction for Australian teenage boys is less than 6 minutes.

Answer 1

t = -4.71

p-value =0.000

Reject H0. We have convincing evidence that the mean time to distraction for Australian teenage boys is less than 6 minutes.

The test performed:

One-Sample t-test

The sample mean is Xˉ=5, the sample standard deviation is s=1.5, and the sample size is n=50. (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: μ =6 Ha: μ <6 This corresponds to a Left-tailed test, for which a t-test for one mean, with unknown population standard deviation will be used. (2a) Critical Value Based on the information provided, the significance level is α=0.01, and the degree of freedom is n-1=50-1=49. Therefore the critical value for this Left-tailed test is tc​=-2.4049. This can be found by either using excel or the t distribution table. (2b) Rejection Region The rejection region for this Left-tailed test is t<-2.4049 (3)Test Statistics The t-statistic is computed as follows: (4) 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 (5) The Decision about the null hypothesis (a) Using traditional method Since it is observed that t=-4.714 < tc​=-2.4049, it is then concluded that the null hypothesis is rejected. (b) Using p-value method Using the P-value approach: The p-value is p=0, and since p=0≤0.01, it is concluded that the null hypothesis is rejected. (6) Conclusion It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population mean μ is less than 6, at the 0.01 significance level.

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