Question

1. A telephone company claims that less than 15% of all college students have their own cell phone plan. A random sample of 70 students revealed that 8 of them had their own plan. Test the company's claim at the 0.05 level of significance.

2. A college statistics instructor claims that the mean age of college statistics students at a local Dallas-based institution is 23. A random sample of 35 college statistics students revealed a mean age of 25.1. The population standard deviation is known to be 4.1 years. Test his claim at the 0.1 level of significance.

3. A random sample of 85 adults ages 18-24 showed that 11 had donated blood within the past year, while a random sample of 254 adults who were at least 25 years old had 18 people who had donated blood within the past year. At the 0.05 level of significance, test the claim that the proportion of blood donors is not equal for these two age groups.

Answer 1

1)

Below are the null and alternative Hypothesis,
Null Hypothesis: p = 0.15
Alternative Hypothesis: p < 0.15

Test statistic,
z = (pcap - p)/sqrt(p*(1-p)/n)
z = (0.1143 - 0.15)/sqrt(0.15*(1-0.15)/70)
z = -0.836

P-value Approach
P-value = 0.2016
As P-value >= 0.05, fail to reject null hypothesis.

b)

Below are the null and alternative Hypothesis,
Null Hypothesis: μ = 23
Alternative Hypothesis: μ ≠ 23

Test statistic,
z = (xbar - mu)/(sigma/sqrt(n))
z = (25.1 - 23)/(4.1/sqrt(35))
z = 3.03

P-value Approach
P-value = 0.0024
As P-value < 0.1, reject the null hypothesis.

c)

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

p1cap = X1/N1 = 11/85 = 0.1294
p1cap = X2/N2 = 18/254 = 0.0709
pcap = (X1 + X2)/(N1 + N2) = (11+18)/(85+254) = 0.0855

Test statistic
z = (p1cap - p2cap)/sqrt(pcap * (1-pcap) * (1/N1 + 1/N2))
z = (0.1294-0.0709)/sqrt(0.0855*(1-0.0855)*(1/85 + 1/254))
z = 1.67

P-value Approach
P-value = 0.095
As P-value >= 0.05, fail to reject null hypothesis.