Question

The business college computing center wants to determine the proportion of business students who have personal computers (PCʹs) at home. If the proportion exceeds 25%, then the lab will scale back a proposed enlargement of its facilities. Suppose 200 business students were randomly sampled and 65 have PCʹs at home. Find the rejection region for this test using α = 0.01. (Note that this is a right-tailed test).
A) Reject H0 if 0 z > 2.33.
B) Reject H0 if 0 z < -2.33.
C) Reject H0 if 0 z > 2.575 or z < -2.575.
D) Reject H0 if 0 z = 2.
Find the test statistic, 0 t , to test the claim about the population mean μ < 6.7 given n = 20, x = 6.3, s = 2.0.
A) -1.233
B) -0.872
C) -1.265
D) -0.894
Determine the test statistic, 0 z , to test the claim about the population proportion p < 0.85 given n = 60 and x = 39.
A) -4.34
B) -1.96
C) -1.85
D) -1.76

Answer 1

Answer:

Question  5

    correct option is  A

Question 6

    correct option is  D

Question 7  

      correct option is  A  

Step-by-step explanation:

Considering question 5

From the question we are told that

     The population proportion considered is  $p = 0.25$

      The sample size is  n  =  200

      The number that had a personal computer at home is  $k = 65$

      The  level of significance is  $\alpha = 0.01$

The null hypothesis is  $H_o : p = 0.25$

The alternative hypothesis is  $H_a : p > 0.25$

Generally from the z-table the critical value of  $\alpha = 0.01$  to the right of the curve is

          $z_{\alpha } = 2.33$

Generally given that it is a right-tailed test , the rejection region is  

      z > 2.33

Considering question 6

  The sample size is  n = 20

  The standard deviation is  $s = 2$

   The sample mean is  $\=x = 6.3$

   The  population mean   $\mu = 6.7$

Generally the test statistics is mathematically represented as

          $t = \frac{ \= x - \mu }{\frac{s}{\sqrt{n} } }$

=>       $t = \frac{ 6.3 - 6.7 }{ \frac{ 2}{ \sqrt{20} } }$

=>       $t = -0.894$

Considering question 7

  The sample size is  n = 60

   The sample mean is  $x = 39$

   The  population proportion    $p = 0.85$

Gnerally the sample proportion is mathematically represented as  

           $\^ p = \frac{x}{n}$

=>       $\^ p = \frac{39}{60}$  

=>       $\^ p = 0.65$

Generally the standard error of this distribution is mathematically represented as

     $SE = \sqrt{ \frac{ p(1 - p ) }{ n } }$

=> $SE = \sqrt{ \frac{ 0.85 (1 - 0.85 ) }{ 60 } }$

=> $SE = 0.0461$

Generally the test statistics is mathematically represented as

          $z = \frac{ \^ p - p }{SE}$

=>       $z = \frac{ 0.65 - 0.85 }{0.0461}$

=>       $z = -4.34$