A manufacturer of nickel-hydrogen batteries randomly selects 100 nickel plates for test cells, cycles them a specified number of times, and determines that 14 of the plates have blistered.(a) Does this provide compelling evidence for concluding that more than 10% of all plates blister under such circumstances? State and test the appropriate hypotheses using a significance level of .05. In reaching your conclusion, what type of error might you have committed?(b) If it is really the case that 15% of all plates blister under these circumstances and a sample size 100 is used, how likely is it that the null hypothesis of part (a) will not be rejected by the 0.05 test? Answer this question for a sample size of 200.(c) How many plates would have to be tested to have \(\beta(.15)\) = 0.10 for the test of part (a)?

Answer:Step-by-step explanation:Test the claim about a single population proportion where the null hypothesis is of the form of {H_0}:p = {p_0}H  The formula for single proportion is,[tex]Z = \frac{{\hat p - p}}{{\sqrt {\frac{{p\left( {1 - p} \right)}}{n}} }}[/tex]​  Here, pp denote the population proportionLet [tex]x[/tex] be the number of successes in a sample of size.Let[tex]n[/tex] be the total sample size.a)Null hypothesis, [tex]H_{0}:p = 0.10[/tex]Alternative hypothesis, [tex]H_{1}:p > 0.10H [/tex]Level of significance, [tex]\alpha = 0.05[/tex]Decision Rule: Reject [tex]H_{0} when p < 0.05p<0.05[/tex]Calculate the test statistic value.[tex]Z = \frac{{\hat p - p}}{{\sqrt {\frac{{p\left( {1 - p} \right)}}{n}} }}\\\\ = \frac{{\left( {\frac{{14}}{{100}}} \right) - 0.10}}{{\sqrt {\frac{{0.10\left( {1 - 0.10} \right)}}{{100}}} }}\\\\ = \frac{{0.14 - 0.10}}{{\sqrt {\frac{{0.10\left( {1 - 0.10} \right)}}{{100}}} }}\\\\ = 1.333[/tex]Calculate the p−value of the test.[tex]p - value = P\left( {Z > {Z_{{\rm{test}}}}} \right)\\\\ = 1 - P\left( {Z \le 1.333} \right)\\\\ = 1 - \left( {{\rm{ = NORMSDIST(1}}{\rm{.333)}}} \right){\rm{ }}\left( {{\rm{Use MS Excel}}} \right)\\\\ = 1 - 0.9087\\\\ = 0.0913[/tex](b)It is really case that 15% of all plates blister under these circumstances and a sample size of 100 is used.[tex]\beta \left( {p'} \right) = \Phi \left[ {\frac{{\left( {{p_0} - p'} \right) + {Z_{{\rm{Critical}}}}\sqrt {\frac{{{p_0}\left( {1 - {p_0}} \right)}}{n}} }}{{\sqrt {\frac{{p'\left( {1 - p'} \right)}}{n}} }}} \right]{\rm{ }}\left( {{\rm{Since}},{Z_{0.05}} = 1.645} \right)\\\\\beta \left( {0.15} \right) = \Phi \left[ {\frac{{\left( {0.10 - 0.15} \right) + 1.645\sqrt {\frac{{0.10\left( {1 - 0.10} \right)}}{{100}}} }}{{\sqrt {\frac{{0.15\left( {1 - 0.15} \right)}}{{100}}} }}} \right]\\\\ = \Phi \left[ {\frac{{ - 0.05 + 0.04935}}{{0.036}}} \right]\\\\ = \Phi \left[ { - 0.02} \right]\\\\ = \left( {{\rm{ = NORMSDIST( - 0}}{\rm{.02)}}} \right){\rm{ }}\left( {{\rm{Use MS Excel function}}} \right)\\\\ = 0.4927[/tex]Hence, the probability that [tex]{H_0}[/tex] will not be rejecting using the [tex]p' = 0.15p[/tex]  is 0.4927.Therefore, there is 49.27% of all samples will result in correct rejection of null hypothesis [tex]\left( {{H_0}} \right)[/tex]For the sample size n = 200,n=200, the probability of type-II error is calculated as follows:[tex]\beta \left( {p'} \right) = \Phi \left[ {\frac{{\left( {{p_0} - p'} \right) + {Z_{{\rm{Critical}}}}\sqrt {\frac{{{p_0}\left( {1 - {p_0}} \right)}}{n}} }}{{\sqrt {\frac{{p'\left( {1 - p'} \right)}}{n}} }}} \right]{\rm{ }}\left( {{\rm{Since}},{Z_{0.05}} = 1.645} \right)\\\\\beta \left( {0.15} \right) = \Phi \left[ {\frac{{\left( {0.10 - 0.15} \right) + 1.645\sqrt {\frac{{0.10\left( {1 - 0.10} \right)}}{{200}}} }}{{\sqrt {\frac{{0.15\left( {1 - 0.15} \right)}}{{200}}} }}} \right]\\\\ = \Phi \left[ { - 0.60} \right]\\\\ = \left( {{\rm{ = NORMSDIST( - 0}}{\rm{.60)}}} \right){\rm{ }}\left( {{\rm{Use MS Excel function}}} \right)\\\\ = 0.2743[/tex](c)The number of plates is,[tex]n = {\left[ {\frac{{{Z_\alpha }\sqrt {{p_0}\left( {1 - {p_0}} \right)} + {Z_\beta }\sqrt {p\left( {1 - p} \right)} }}{{p - {p_0}}}} \right]^2}\\\\ = {\left[ {\frac{{1.645\sqrt {0.1\left( {1 - 0.1} \right)} + 1.28\sqrt {0.15\left( {1 - 0.15} \right)} }}{{0.15 - 0.10}}} \right]^2}\\\\ = {\left( {19.01} \right)^2}\\\\ = 362[/tex]