Question

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 =

Answer 1

Answer:

(a) Null Hypothesis, $H_0$ : p = 71%   

    Alternate Hypothesis, $H_A$ : p $\neq$ 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, $H_0$ : p = 71%   {means that % of college students who work is same as 71% since 2011}

Alternate Hypothesis, $H_A$ : p $\neq$ 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.  =  $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

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

           n = sample of students = 110

(b) So, test statistics  =  $\frac{\frac{91}{110}-0.71}{\sqrt{\frac{\frac{91}{110}(1-\frac{91}{110})}{110} } }$

                                    =  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 $\leq$ 3.25)

                                            = 1 - 0.99942 = 0.0006

So, the p-value is 0.0006.