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The attention span of children (ages 3 to 5) is claimed to be Normally distributed with a mean of 15 minutes and a standard deviation of 4 minutes. A test is to be performed to decide if the average attention span of these kids is really this short or if it is longer. You decide to test the hypotheses H0: μ = 15 versus Ha: μ > 15 at the 5% significance level. A sample of 10 children will watch a TV show they have never seen before, and the time until they walk away from the show will be recorded. At a significance level of 5%, the decision rule would be to reject the null hypothesis if the observed sample mean is greater than __________________ minutes.

Answer :

Answer:

If the observed sample mean is greater than 17.07 minutes, then we would reject the null hypothesis.

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 15 minutes

Sample size, n = 10

Alpha, α = 0.05

Population standard deviation, σ = 4 minutes

First, we design the null and the alternate hypothesis

[tex]H_{0}: \mu = 15\text{ minutes}\\H_A: \mu > 15\text{ minutes}[/tex]

Since the population standard deviation is given, we use one-tailed z test to perform this hypothesis.

Formula:

[tex]z_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }[/tex]

Now, [tex]z_{critical} \text{ at 0.05 level of significance for one tail} = 1.64[/tex]

Thus, we would reject the null hypothesis if the z-statistic is greater than this critical value.

Thus, we can write:

[tex]z_{stat} = \displaystyle\frac{\bar{x} - 15}{\frac{4}{\sqrt{10}} } > 1.64\\\\\bar{x} - 15>1.64\times \frac{4}{\sqrt{10}}\\\bar{x} - 15>2.07\\\bar{x} > 15+2.07\\\bar{x} > 17.07[/tex]

Thus, the decision rule would be if the observed sample mean is greater than 17.07 minutes, then we would reject the null hypothesis.

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