The Parktown prawn is an impressively repellent large insect, common in Johannesburg (look them up on the Web). I claim that their average length is 10 cm. You collect 100 Parktown prawns (this will take about 10 mins, in the right places in Johannesburg; more difficult from the US). The mean length of these prawns is 7 cm. The standard deviation is 1 cm. Assess the evidence against my claim.

Respuesta :

Answer:

There is enough statistical evidence to claim that the average length is not 10 cm.

Step-by-step explanation:

The claim that the average lenght is 10 cm, is the null hypothesis (has the "equal" sign). The claim that the average length is not 10 cm is the alternative hypothesis.

[tex]H_0:\mu=10\\\\H_a:\mu\neq 10[/tex]

The significance level is assumed to be 0.10.

The sample mean is 7 cm and the sample standard deviation is 1 cm.

The degrees of freedom are

[tex]df=n-1=100-1=99[/tex]

The t-statistic is:

[tex]t=\frac{M-\mu}{\sigma/\sqrt{n}} =\frac{7-10}{1/\sqrt{100}}=\frac{-3}{0.1}=-30[/tex]

The P-value of t=-30 is P=0.

This means there is almost no chance of getting this sample if the mean is really 10 cm.

There is enough statistical evidence to claim that the average length is not 10 cm.

The required average value is 10.

Critical value

The critical value in statistics is the measurement statisticians use to quantify the margin of error within a collection of data.

Given that,

[tex]H_0:\mu=10\\H_1:\mu \ne10 \ for\ 0.05\\critical\ value=\pm1.96\\z=\frac{7-10}{1}\\ z=-3[/tex]

Since [tex]z[/tex] is greater than the critical value, then we reject [tex]H_0[/tex] and say that there are such evidence to reject that average is 10.

Learn more about the topic Critical value:

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