Question

Based on information from a large insurance company, 67% of all damage liability claims are made by single people under the age of 25. A random sample of 54 claims showed that 44 were made by single people under the age of 25. Does this indicate that the insurance claims of single people under the age of 25 is higher than the national percent reported by the large insurance company? State the null and alternate hypothesis.

a) H_o: p = .67, H_a: p < .67

b) H_o: p = .67, H_a: p ≠ .67

c) H_o: p = .81, H_a: p > .81

d) H_o: p = .67, H_a: p > .67

e) H_o: p = .81, H_a: p < .81

Based on information from a large insurance company, 66% of all damage liability claims are made by single people under the age of 25. A random sample of 53 claims showed that 43 were made by single people under the age of 25. Does this indicate that the insurance claims of single people under the age of 25 is higher than the national percent reported by the large insurance company? Give the test statistic and your conclusion.

a) z = -2.326; reject H_o at the 5% significance level

b) z = 1.826; reject H_o at the 5% significance level

c) z = 2.326; reject H_o at the 5% significance level

d) z = 2.326; fail to reject H_o at the 5% significance level

e) z = -1.826; fail to reject H_o at the 5% significance level

Answer 1

Solution:

For your first question:

Given:

67% of all damage liability claims are made by single people under the age of 25.

Claim: Insurance claims of single people under the age of 25 is higher than the national percent reported by the large insurance company

That means population proportion is higher than 67%.

This can be write down in symbolic form as,

Null and alternative hypothesis:

Ho: p = .67, Ha: p > .67

Therefore, option (d) is correct

For your second question:

Given:

Given:

Sample size = n = 53

x= 43

Sample proportion:

This corresponding to a right-tailed test, for which a z-test for one population proportion needs to be used.

Test statistic:

Test statistic z = 2.326

Using the P-value approach:

P(Z>2.326) = 1-P(Z<2.326) = 0.01

The p-value is p = 0.01 …Using standard normal table

And α = 0.05

Since p = 0.01 <0.05, it is concluded that reject Ho at the 5% significance level. (Option (c) is correct)

Done