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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

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