Question

On average, Americans have lived in 4 places by the time they are 18 years old. Is this average a different number for college students? The 50 randomly selected college students who answered the survey question had lived in an average of 3.79 places by the time they were 18 years old. The standard deviation for the survey group was 1.2. What can be concluded at the α = 0.10 level of significance?

Answer 1

Answer:

The decision rule is

Fail to reject the null hypothesis

The conclusion is

There is  no sufficient evidence to conclude that the average is a different number for college students

Step-by-step explanation:

From the question we are told that

    The  sample size is  n =  50  

    The  population mean is $\mu = 4$

     The  sample mean is $\= x = 3.79$

     The standard deviation is $\sigma = 1.2$

     The  level of significance is  $\alpha = 0.10$

The  null hypothesis is  $H_o : \mu = 4$

The alternative hypothesis is $H_a : \mu \ne 4$

Generally  the  test statistics is mathematically represented as

         $z = \frac{ \= x - \mu }{ \frac{ \sigma }{ \sqrt{n}} }$

=>      $z = \frac{ 3.79 - 4 }{ \frac{1.2}{ \sqrt{50}} }$

=>       $z = -1.237$

From the z table  the area under the normal curve to the left corresponding to    -1.237  is    

         $P(Z < -1.237) = 0.10804$

Generally the p-value is mathematically represented as

        $p-value = P(Z < -1.237) = 2 * 0.10804$

=>    $p-value = 0.21608$

Generally  from the values obtained we see that

       $p-value > \alpha$

Hence

The decision rule is

Fail to reject the null hypothesis

 The conclusion is

There is  no sufficient evidence to conclude that the average is a different number for college students