Answered

Westonci.ca makes finding answers easy, with a community of experts ready to provide you with the information you seek. Discover a wealth of knowledge from experts across different disciplines on our comprehensive Q&A platform. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly platform.

Childhood participation in sports, cultural groups, and youth groups appears to be related to improved self-esteem for adolescents (McGee, Williams, Howden-Chapman, Martin, & Kawachi, 2006). In a representative study, a sample of n = 100 adolescents with a history of group participation is given a standardized self-esteem questionnaire. For the general population of adolescents, scores on this questionnaire form a normal distribution with a mean of μ = 50 and a standard deviation of σ = 15. The sample of group-participation adolescents had an average of M = 53. 8.

Does this sample provide enough evidence to conclude that self-esteem scores for these adolescents are significantly different from those of the general population? Use a two-tailed test with α =. 5.

The null hypothesis is H₀: , even group participation.

The standard error is , so z =

(to two decimal places). Using the Distributions tool, the critical value of z =

Sagot :

Using the z-distribution, as we have the standard deviation for the population, it is found that the sample provides enough evidence to conclude that self-esteem scores for these adolescents are significantly different from those of the general population.

What are the hypothesis tested?

At the null hypothesis, we test if the score for group-participation adolescents is the same as the general population, that is:

[tex]H_0: \mu = 50[/tex]

At the alternative hypothesis, we test if the score is different, that is:

[tex]H_1: \mu \neq 50[/tex]

What is the test statistic?

The test statistic is given by:

[tex]z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]

The parameters are:

  • [tex]\overline{x}[/tex] is the sample mean.
  • [tex]\mu[/tex] is the value tested at the null hypothesis.
  • [tex]\sigma[/tex] is the standard deviation of the population.
  • n is the sample size.

In this problem, we have that:

[tex]\overline{x} = 53.8, \mu = 50, \sigma = 15, n = 100[/tex].

Hence:

[tex]z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]

[tex]z = \frac{53.8 - 50}{\frac{15}{\sqrt{100}}}[/tex]

[tex]z = 2.53[/tex]

What is the decision?

Considering a two-tailed test, as we are testing if the mean is different of a value, with a significance level of 0.05, the critical value is [tex]|z^{\ast}| = 1.96[/tex]

Since the absolute value of the test statistic is greater than the critical value for the two-tailed test, it is found that the sample provides enough evidence to conclude that self-esteem scores for these adolescents are significantly different from those of the general population.

More can be learned about the z-distribution at https://brainly.com/question/26454209

Thanks for stopping by. We are committed to providing the best answers for all your questions. See you again soon. Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. Stay curious and keep coming back to Westonci.ca for answers to all your burning questions.