Question

1. Walking to improve health. In a study investigating a link between walking and improved health (Social Science & Medicine, Apr. 2014), researchers reported that adults walked an average of 5.5 days in the past month for the purpose of health or recreation. a) Specify the null and alternative hypotheses for testing whether the true average number of days in the past month that adults walked for the purpose of health or recreation is lower than 5.5 days. b) Is the test one-tailed or two-tailed? c) Using α = .01, specify the critical value (CV), the rejection region, and draw the distribution.

Answer 1

a) We need to conduct a hypothesis in order to check if the true average number of days in the past month that adults walked for the purpose of health or recreation is lower than 5.5 days (a;ternative hypothesis) , the system of hypothesis would be:

Null hypothesis:
`\mu \geq 5.5`

Alternative hypothesis:
`\mu < 5.5`

b) For this case since we want to determine if the true mean is lower than an specified value is a one tailed test

c) For this case the significance level is given
`\alpha=0.01` and we want to find the critical value, so we need to find a critical value on the normal standard distribution who accumulate 0.01 of the area on the right tail and if we use excel or a table we got:

`z_(cric)= -2.326`

And the rejection zone would be:
`(\infty ,-2.326)`

And the region is on the figure attached

Explanation:

Part a: State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the true average number of days in the past month that adults walked for the purpose of health or recreation is lower than 5.5 days (a;ternative hypothesis) , the system of hypothesis would be:

Null hypothesis:
`\mu \geq 5.5`

Alternative hypothesis:
`\mu < 5.5`

Part b

For this case since we want to determine if the true mean is lower than an specified value is a one tailed test

Part c

For this case the significance level is given
`\alpha=0.01`and we want to find the critical value, so we need to find a critical value on the normal standard distribution who accumulate 0.01 of the area on the right tail and if we use excel or a table we got:

`z_(cric)= -2.326`

And the rejection zone would be:
`(\infty ,-2.326)`

And the region is on the figure attached