Bill Alther is a zoologist who studies Anna's hummingbird (Calypte anna).† Suppose that in a remote part of the Grand Canyon, a random sample of six of these birds was caught, weighed, and released. The weights (in grams) were as follows. 3.7 2.9 3.8 4.2 4.8 3.1 The sample mean is x = 3.75 grams. Let x be a random variable representing weights of hummingbirds in this part of the Grand Canyon. We assume that x has a normal distribution and σ = 0.86 gram. Suppose it is known that for the population of all Anna's hummingbirds, the mean weight is μ = 4.50 grams. Do the data indicate that the mean weight of these birds in this part of the Grand Canyon is less than 4.50 grams? Use α = 0.10.

If we compare the p value and a significance level assumed for example [tex]\alpha=0.05[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we reject the null hypothesis, and the actual true mean for the weights is significant less than 4.5 grams.

Step-by-step explanation:

Data given and notation

[tex]\bar X=3.75[/tex] represent the sample mean

[tex]\sigma=0.86[/tex] represent the standard deviation for the population

[tex]n=6[/tex] sample size

[tex]\mu_o =4.5[/tex] represent the value that we want to test

[tex]\alpha=0.1[/tex] represent the significance level for the hypothesis test.

z would represent the statistic (variable of interest)

[tex]p_v[/tex] represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to determine if the mean is less than 4.5 grams, the system of hypothesis would be:

Null hypothesis:[tex]\mu \geq 4.5[/tex]

Alternative hypothesis:[tex]\mu < 4.5[/tex]

We know the population deviation, so for this case is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:

[tex]z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}[/tex] (1)

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

[tex]z=\frac{3.75-4.5}{\frac{0.86}{\sqrt{6}}}=-2.136[/tex]

Calculate the P-value

Since is a one-side left tailed test the p value would be:

[tex]p_v =P(Z<-2.136)=0.016[/tex]

Conclusion

If we compare the p value and a significance level assumed for example [tex]\alpha=0.05[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we reject the null hypothesis, and the actual true mean for the scores is significant less than 4.5grams.