For a study of adolescents' disclosure of their dating and romantic relationships, a sample of 224 high school students was surveyed. One of the variables of interest was the level of disclosure to an adolescent's mother (measured on a 5-point scale, where 1 = "never tell" and 5 ="always tell"). The sampled high school students had a mean disclosure score of 3.38 and a standard deviation of 0.98. The researchers hypothesize that the true mean disclosure score of all adolescents will exceed 3. Do you believe the researchers?

Conduct a formal test of hypothesis using α = 0.05. Set

Respuesta :

Answer:

 The  decision rule is  

Reject the null hypothesis

   The conclusion is

There sufficient evidence that the true mean disclosure score of all adolescents will exceed 3

Step-by-step explanation:

From the question we are told that  

      The sample is  n  =  224

      The  sample mean is [tex]\= x = 3.38[/tex]

      The standard deviation is  [tex]s = 0.98[/tex]

      The population mean is [tex]\mu = 3[/tex]

       The level of significance  is  [tex]\alpha = 0.05[/tex]

The null hypothesis is  [tex]H_o : \mu = 3[/tex]

The alternative hypothesis is  [tex]H_a : \mu > 3[/tex]

Generally the test statistics is mathematically represented as

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

=>       [tex]z = \frac{ 3.38 - 3 }{ \frac{0.98 }{ \sqrt{224} } }[/tex]

=>       [tex]z = 5.80[/tex]

From the z table  the area under the normal curve to the right corresponding to 5.80  is

        [tex]p-value = 0[/tex]

From the value obtained we that [tex]p-value < \alpha[/tex] , hence  

   The  decision rule is  

Reject the null hypothesis

   The conclusion is

There sufficient evidence that the true mean disclosure score of all adolescents will exceed 3