Question

Consider the following hypotheses: H0: p ? 0.38 HA: p < 0.38

Compute the p-value based on the following sample information

a. x = 22; n = 74

b. x = 110; n = 300

c. pbar = 0.34; n = 50

d. pbar = 0.34; n = 400

Answer 1

Consider the following hypotheses: H0: p ? 0.38 HA: p < 0.38

To find the p-values for following cases:

a) x = 22; n = 74

$\hat p =\frac{x}{n}=\frac {22}{74}=0.2973$

Since given test is left tailed test,

$Pvalue=P(\hat p < 0.2973)$

              $=P(\frac{\hat p - p}{\sqrt {\frac {p(1-p)}{n}}}< \frac{0.2973 - 0.38}{\sqrt {\frac {0.38(1-0.38)}{74}}})$

             

             

              $=1-0.9292$

              $=0.0708$

Hence, P-value = 0.0708

                  

b) x = 110; n = 300

$\hat p =\frac{x}{n}=\frac {110}{300}=0.3667$

Since given test is left tailed test,

$Pvalue=P(\hat p < 0.3667)$

              $=P(\frac{\hat p - p}{\sqrt {\frac {p(1-p)}{n}}}< \frac{0.3667 - 0.38}{\sqrt {\frac {0.38(1-0.38)}{300}}})$

             

             

              $=1-0.6808$

              $=0.3192$

Hence, P-value = 0.3192

c) pbar = 0.34; n = 50

Since given test is left tailed test,

$Pvalue=P(\hat p < 0.34)$

              $=P(\frac{\hat p - p}{\sqrt {\frac {p(1-p)}{n}}}< \frac{0.34 - 0.38}{\sqrt {\frac {0.38(1-0.38)}{50}}})$

             

             

              $=1-0.7190$

              $=0.2810$

Hence, P-value = 0.2810

d) pbar = 0.34; n = 400

Since given test is left tailed test,

$Pvalue=P(\hat p < 0.34)$

              $=P(\frac{\hat p - p}{\sqrt {\frac {p(1-p)}{n}}}< \frac{0.34 - 0.38}{\sqrt {\frac {0.38(1-0.38)}{400}}})$

             

             

              $=1-0.9505$

              $=0.0495$

Hence, P-value = 0.0495