Question

A study is conducted regarding shatterproof glass used in automobiles. Twenty-six glass panes are coated with an anti-shattering film. Then a 5-pound metal ball is fired at 70mph at each pane. Five of the panes shatter. We wish to determine whether, in the population of all such panes, the probability the glass shatters under these conditions is different from π= 0.2

(a) State the appropriate null and alternative hypotheses.

(b) Check the conditions for trusting the conclusion of the test, and calculate the observed value of an appropriate test statistic.

(c) Calculate the rejection region and draw a conclusion, given the significance level α= 0.05.

(d) Calculate the p-value.

(e) Compute the power of the test if the trueπwas in fact 0.3.

Answer 1

a)

null Hypothesis:               Ho:		p	=	0.200
alternate Hypothesis:    Ha:		p	≠	0.200

b) for np and n(1-p) both are greater than or equal to 5 and sample is random, therefore condition is satisfied for normal approximation of binomial distribution

sample success                                                           x                    =				5
sample size		n                    =		26
std error       =Se		=√(p*(1-p)/n) =		0.0784
sample proportion     p̂		x/n=		0.1923
test stat = z                =		(p̂-p)/Se=		-0.10

c)

Decision rule :                   reject Ho if absolute value of test statistic |z|>1.96

as test statistic does not  falls in rejection region we can not reject null . We do not have evidence to conclude that probability the glass shatters under these conditions is different from π= 0.2

d)

p value =0.9218

e)