Question

Assume that a random sample of 4 Goldfinch birds are going to be used to test the above hypothesis. Compute the largest value of the sample mean, that will allow the experimenter reject the null and prove the alternative at significance level 0.01. For this problem, make sure that your solution includes all steps of your computation. You will not earn points if you just use a formula

Answer 1

This question is incomplete, the complete question is;

Assume that body masses of Goldfinch birds follow a normal distribution, with a standard deviation equal to 0.04 oz. An ornithologist who would like to make some inference about the average body mass of the Goldfinch birds. In particular, at a significance level of 0.01, she would like to test the null hypothesis H₀: Average body mass of the Goldfinch birds is 0.5 oz, against the alternative claim that average body size is less than 0.5 oz.

      Assume that a random sample of 4 Goldfinch birds are going to be used to test the above hypothesis. Compute the largest value of the sample mean, that will allow the experimenter reject the null and prove the alternative at significance level 0.01. For this problem, make sure that your solution includes all steps of your computation. You will not earn points if you just use a formula

Answer: the largest value of the sample mean that will enable the experimenter to reject the null hypothesis is 0.4534 oz.

Step-by-step explanation:

Let us consider testing a hypothesis about a population mean with population standard deviation that is known,

Let the average body mass of the Goldfinch bird be μ^μ

Now our hypotheses are will be;

H₀ : μ = 0.5    and

H₁ : μ < 0.5

For the test, we shall use the z statistic which will be

z = (x"-μ)/(σ/√n)

Now given that;

σ = 0.04 and n = 4  

sample mean x" = ?

Now in order to be able to reject the null hypothesis, the p-value should be ≤ α = 0.01.

so to find the largest value of sample mean, we will use the maximum of p-value that is 0.01.

For a p-value = 0.01 at the lower tail (which is the percentile for z),

the corresponding z will be -2.33

so we substitute our values into our formula

z = (x"-μ)/(σ/√n)

z(σ/√n) = (x"-μ)

-2.33(0.04/√n) = (x"-0.5)

-0.0466 = x" - 0.5

x" = 0.5 - 0.0466

x"= 0.4534

therefore the largest value of the sample mean that will enable the experimenter to reject the null hypothesis is 0.4534 oz.