Question

A simple random sample of size n = 200 drivers were asked if they drive a car manufactured in a certain country. Of the 200 drivers​ surveyed, 110 responded that they did. Determine if more than half of all drivers drive a car made in this country at the alpha equals 0.05 level of significance.
Determine the null and alternative hypotheses.

Answer 1

Answer:

Null hypothesis: H0 = 0.50

Alternative hypothesis: Ha > 0.50

z = 1.414

P value = P(Z<-3.16) = 0.0787

Decision: not enough evidence to reject the null hypothesis. Accept null hypothesis.

Not more than half of all drivers drive a car made in this country

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

Step-by-step explanation:

Given;

n= 200 represent the random sample taken

Null hypothesis: H0 = 0.50

Alternative hypothesis: Ha > 0.50

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 = 200

po = Null hypothesized value = 0.50

p^ = Observed proportion = 110/200 = 0.55

Substituting the values we have

z = (0.55-0.50)/√{0.50(1-0.50)/200}

z = 1.414

z = 1.414

To determine the p value (test statistic) at 0.05 significance level, using a one tailed hypothesis.

P value = P(Z>1.414) = 0.0787

Since z at 0.05 significance level is between -1.96 and +1.96 and the z score for the test (z = 1.414) which falls within the region bounded by Z at 0.05 significance level. And also the one-tailed hypothesis P-value is 0.0787 which is greater 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.