Question

Wyoming fisheries contend that the mean number of cutthroat trout caught during a full day of fly-fishing on the Snake, Buffalo, and other rivers and streams in the Jackson Hole area is 4.0. To make their yearly update, the fishery personal asked a sample of fly-fishermen to keep a count of the number caught during the day. The numbers were: 4, 4, 3, 2, 6, 8, 7, 1,9, 3, 1, and 6. At the 0.05 significance level, can we conclude that the mean number caught is greater than 4.0?​

Answer 1

`t=(4.5-4)/((2.680)/(√(10)))=0.59`

The degrees of freedom are given by:

`df =n-1= 12-1=11`

And the p value would be:

`p_v =2*P(t_(11)>0.59)=0.567`

Since the p value is higher than the significance level we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true mean area is not significantly different from 4

Explanation:

We have the following data given:

4, 4, 3, 2, 6, 8, 7, 1,9, 3, 1, and 6

The sample mean and deviation from these data are:

`\bar X=4.5` represent the sample mean

`s=2.680` represent the sample deviation

`n=10` sample size

`\mu_o =4` represent the value to verify

`\alpha=0.05` represent the significance level

t would represent the statistic

`p_v` represent the p value

Hypothesis to verify

We want to verify if the true mean is equal to 4, the system of hypothesis would be:

Null hypothesis:
`\mu =4`

Alternative hypothesis:
`\mu \\eq 4`

The statistic is given by:

`t=(\bar X-\mu_o)/((s)/(√(n)))` (1)

Replacing the the info we got:

`t=(4.5-4)/((2.680)/(√(10)))=0.59`

The degrees of freedom are given by:

`df =n-1= 12-1=11`

And the p value would be:

`p_v =2*P(t_(11)>0.59)=0.567`

Since the p value is higher than the significance level we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true mean area is not significantly different from 4