Question

A survey of 1060people who took trips revealed that 94 of them included a visit to a theme park. Based on those survey results, a management consultant claims that less than 11 % of trips include a theme park visit. Test this claim using the ?=0.01significance level.

(a) The test statistic is ___

(b) The P-value is ___

(c) The conclusion is  

A. There is sufficient evidence to support the claim that less than 11 % of trips include a theme park visit.
B. There is not sufficient evidence to support the claim that less than 11 % of trips include a theme park visit.

Independent random samples, each containing 90 observations, were selected from two populations. The samples from populations 1 and 2 produced 36 and 26 successes, respectively.
Test ?0:(?1−?2)=0against ??:(?1−?2)>0 Use ?=0.1

(a) The test statistic is ___

(b) The P-value is ___

(c) The final conclusion is
A. There is not sufficient evidence to reject the null hypothesis that (?1−?2)=0
B. We can reject the null hypothesis that (?1−?2)=0 and conclude that (?1−?2)>0

Answer 1

1)

Below are the null and alternative Hypothesis,
Null Hypothesis:H0 : p = 0.11
Alternative Hypothesis:HA: p < 0.11

Test statistic,
z = (pcap - p)/sqrt(p*(1-p)/n)
z = (0.0887 - 0.11)/sqrt(0.11*(1-0.11)/1064)
z = -2.22

P-value Approach
P-value = 0.0132
As P-value >= 0.01, fail to reject null hypothesis.

B. There is not sufficient evidence to support the claim that less than 11 % of trips include a theme park visit.

2)

Below are the null and alternative Hypothesis,
Null Hypothesis, H0: p1 = p2
Alternate Hypothesis, Ha: p1 > p2

p1cap = X1/N1 = 36/90 = 0.4
p1cap = X2/N2 = 26/90 = 0.2889
pcap = (X1 + X2)/(N1 + N2) = (36+26)/(90+90) = 0.3444

Test statistic
z = (p1cap - p2cap)/sqrt(pcap * (1-pcap) * (1/N1 + 1/N2))
z = (0.4-0.2889)/sqrt(0.3444*(1-0.3444)*(1/90 + 1/90))
z = 1.57

P-value Approach
P-value = 0.0584
As P-value < 0.1, reject the null hypothesis.

B. We can reject the null hypothesis that (?1−?2)=0 and conclude that (?1−?2)>0