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In 2011, a U.S. Census report determined that 71% of college students work. A researcher thinks this percentage has changed since then. A survey of 110 college students reported that 91 of them work. Is there evidence to support the reasearcher's claim at the 1% significance level? A normal probability plot indicates that the population is normally distributed.

a) Determine the null and alternative hypotheses.

H0: p =
Ha:P Select an answer (Put in the correct symbol and value)

b) Determine the test statistic. Round to two decimals.
c) Find the p-value. Round to 4 decimals.

P-value =

Respuesta :

Answer:

(a) Null Hypothesis, [tex]H_0[/tex] : p = 71%   

    Alternate Hypothesis, [tex]H_A[/tex] : p [tex]\neq[/tex] 71%  

(b) The test statistics is 3.25.

(c) The p-value is 0.0006.

Step-by-step explanation:

We are given that a U.S. Census report determined that 71% of college students work. A researcher thinks this percentage has changed since then.

A survey of 110 college students reported that 91 of them work.

Let p = proportion of college students who work

(a) Null Hypothesis, [tex]H_0[/tex] : p = 71%   {means that % of college students who work is same as 71% since 2011}

Alternate Hypothesis, [tex]H_A[/tex] : p [tex]\neq[/tex] 71%   {means that % of college students who work is different from 71% since 2011}

The test statistics that will be used here is One-sample z proportion statistics;

                                 T.S.  =  [tex]\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] ~ N(0,1)

where, [tex]\hat p[/tex] = sample proportion of college students who reported they work = [tex]\frac{91}{110}[/tex] = 82.73%

           n = sample of students = 110

(b) So, test statistics  =  [tex]\frac{\frac{91}{110}-0.71}{\sqrt{\frac{\frac{91}{110}(1-\frac{91}{110})}{110} } }[/tex]

                                    =  3.25

The test statistics is 3.25.

(c) P-value of the test statistics is given by the following formula;

       P-value = P(Z > 3.25) = 1 - P(Z [tex]\leq[/tex] 3.25)

                                            = 1 - 0.99942 = 0.0006

So, the p-value is 0.0006.

The null and alternative hypotheses is [tex]\rm H_0:[/tex] p = 71% and [tex]\rm H_a[/tex] : p [tex]\neq[/tex] 71%, the value test statistics is 3.25, and the p-value is 0.0006.and this can be determined by using the given data.

Given :

  • In 2011, a U.S. Census report determined that 71% of college students work.
  • A survey of 110 college students reported that 91 of them work.

a) The hypothesis is given by:

Null hypothesis --   [tex]\rm H_0:[/tex] p = 71%

Alternate Hypothesis --  [tex]\rm H_a[/tex] : p [tex]\neq[/tex] 71%

b) The statistics test is given by:

[tex]\rm TS = \dfrac{\hat{p}-p}{\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}} }[/tex]

[tex]\rm TS = \dfrac{\dfrac{91}{110}-0.71}{\sqrt{\dfrac{\dfrac{91}{110}(1-\dfrac{91}{110})}{110}} }[/tex]

Simplify the above expression.

TS = 3.25

c) The p-value is given by:

P-value = P(Z>3.25) = 1 - P(Z [tex]\leq[/tex] 3.25)

                                = 1 - 0.99942

                                = 0.0006

For more information, refer to the link given below:

https://brainly.com/question/17716064