Question

Suppose Zanna is a biologist who studies meerkats in Botswana. While conducting research for her PhD, she found that the average weight of adult female meerkats born in zoos worldwide is 723.53 g723.53 g with a standard deviation of σ=125.08 g.σ=125.08 g. She wants to find out if adult female meerkats that were born in the spring are heavier than the other meerkats. She decides to carry out a one‑sample z-testz-test for a mean with a significance level of α=0.01.α=0.01. She finds the average weight of a random sample of n=888n=888 meerkats born in the spring to be ¯¯¯x=734.42 g.

What are Zanna’s null (H0H0) and alternative (H1H1) hypotheses?

a. H0:μ=734.42 g, H1:μ>734.42 gH0:μ=734.42 g, H1:μ>734.42 g

b. H0:μ=723.53 g, H1:μ>723.53 gH0:μ=723.53 g, H1:μ>723.53 g

c. H0:μ>734.42 g, H1:μ=734.42 gH0:μ>734.42 g, H1:μ=734.42 g

d. H0:μ=723.53 g, H1:μ≠723.53 gH0:μ=723.53 g, H1:μ≠723.53 g

What is the value of the zz-statistic for Zanna's test? Please round your answer to two decimal places.

z=z=

What is the critical zz‑value for a significance level of α=0.01α=0.01? Please round your answer to two decimal places.

critical z=z=

Based on her results, Zanna should

a. fail to reject the null. There is insufficient evidence (P>0.01P>0.01) that the mean weight of meerkats born in the spring is greater than the population mean of 723.53 g.

b. fail to reject the null. There is sufficient evidence (P>0.01P>0.01) that the mean weight of meerkats born in the spring is greater than the population mean of 723.53 g.

c. reject the null. There is insufficient evidence (P<0.01P<0.01) that the mean weight of meerkats born in the spring is greater than the population mean of 723.53 g.

d. reject the null. There is sufficient evidence (P<0.01P<0.01) that the mean weight of meerkats born in the spring is greater than the population mean of 723.53 g.

e. reject the null. There is sufficient evidence (P<0.01P<0.01) that the mean weight of meerkats born in the spring is greater than the sample mean of 734.42 g.

Answer 1

This is the right tailed test .

The null and alternative hypothesis is

H₀ : = 723.53

H_a : > 723.53

Test statistic = z

= ( - ) /    / n

= (734.42 - 23.53) / 125.08 / 888

Test statistic = 2.59

Critical value = 2.33  

P-value <

Reject the null hypothesis .

d. reject the null. There is sufficient evidence (P<0.01) that the mean weight of meer kats born in the spring is greater than the population mean of 723.53 g.

There is sufficient evidence to conclude that the