Question

UCF believes that the average time someone spends in the gym is 56 minutes. The university statistician takes a random sample of 32 gym goers and finds the average time of the sample was 50 minutes. Assume it is known the standard deviation of time all people spend in the gym is 8 minutes. What conclusion can the university statistician make?

Answer 1

We conclude that the average time someone spends in the gym is different from 56 minutes.

Explanation:

We are given that UCF believes that the average time someone spends in the gym is 56 minutes.

The university statistician takes a random sample of 32 gym goers and finds the average time of the sample was 50 minutes. Assume it is known the standard deviation of time all people spend in the gym is 8 minutes.

Let
`\mu` = population average time someone spends in the gym

SO, Null Hypothesis,
`H_0` :
`\mu` = 56 minutes {means that the average time someone spends in the gym is 56 minutes}

Alternate Hypothesis,
`H_A` :
`\mu\\eq` 56 minutes {means that the average time someone spends in the gym is different from 56 minutes}

The test statistics that will be used here is One-sample z test statistics as we know about the population standard deviation;

T.S. =
`(\bar X -\mu)/((\sigma)/(√(n) ) )` ~ N(0,1)

where,
`\bar X` = sample average time someone takes in the gym = 50 min

`\sigma` = population standard deviation = 8 minutes

n = sample of gym goers = 32

So, test statistics =
`(50-56)/((8)/(√(32) ) )`

= -4.243

Since in the question we are not given the level of significance so we assume it to b 5%. Now at 5% significance level, the z table gives critical value between -1.96 and 1.96 for two-tailed test. Since our test statistics does not lie within the range of critical values of z so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region.

Therefore, we conclude that the average time someone spends in the gym is different from 56 minutes.