Answer:
Null hypothesis: the proportion of customers that were able to program their DVD player is equal to 0.30
H0: p = 0.30
Alternative hypothesis: the proportion of customers that were able to program their DVD player is less than 0.30
Ha: p < 0.30
z = -0.558
P value = P(Z<-0.558) = 0.288
Decision we fail to reject the null hypothesis, so we'll accept the Null hypothesis.
Rule
If;
P-value > significance level --- accept Null hypothesis
P-value < significance level --- reject Null hypothesis
Z score > Z(at 95% confidence interval) ---- reject Null hypothesis
Z score < Z(at 95% confidence interval) ------ accept Null hypothesis
Explanation:
Given;
n = 60 represent the random sample taken
Null hypothesis: the proportion of customers that were able to program their DVD player is equal to 0.30
H0: p = 0.30
Alternative hypothesis: the proportion of customers that were able to program their DVD player is less than 0.30
Ha: p < 0.30
Test statistic z score can be calculated with the formula below;
z = (p^−po)/√{po(1−po)/n}
Where,
z= Test statistics
n = Sample size = 60
po = Null hypothesized value = 0.30
p^ = Observed proportion = 16/60 = 0.267
Substituting the values we have
z = (0.267-0.30)/√(0.30(1-0.30)/60)
z = -0.5578
z = -0.558
To determine the p value (test statistic) at 0.05 significance level, using a one tailed hypothesis.
P value = P(Z<-0.558) = 0.28774 = 0.288
Since z at 0.05 significance level is between -1.96 and +1.96 and the z score for the test (z = -0.558) which falls with the region bounded by Z at 0.05 significance level. And also the one-tailed hypothesis P-value is 0.288 which is higher than 0.05. Then we can conclude that we don't have enough evidence to FAIL or reject the null hypothesis, and we can say that at 5% significance level the null hypothesis is valid, therefore we accept the Null hypothesis