QUESTION 6 A marketing research firm wishes to estimate the

QUESTION 6.

A marketing research firm wishes to estimate the percentage of homeowners who are dissatisfied with their present homeowner’s insurance policy. A simple random sample of 400 homeowners led to 80 who were dissatisfied with their homeowner’s insurance policy.

a. Compute the 95% confidence interval for the population percentage of homeowners who are dissatisfied with their present homeowner’s insurance policy.
b. Interpret this confidence interval.
c. How many homeowners should be sampled in order to be 95% confident of being within 2% of the population percentage of homeowners who are dissatisfied with their present homeowner’s insurance policy?

Question 7.

Persons living near a smelting plant have complained that the plant violates the city’s noise pollution code. The code states that to be in compliance, noise levels are only allowed to exceed 120 decibels less than 10% of the time. You monitor the noise levels at 150 randomly selected times and found that 11 were above 120 decibels. Does the sample data provide evidence to conclude that the plant is in compliance with the noise pollution code (witha = .05)? Use the hypothesis testing procedure outlined below.

a. Formulate the null and alternative hypotheses.
b. State the level of significance.
c. Find the critical value (or values), and clearly show the rejection and nonrejection regions.
d. Compute the test statistic.
e. Decide whether you can reject Ho and accept Ha or not.
f. Explain and interpret your conclusion in part e. What does this mean?
g. Determine the observed p-value for the hypothesis test and interpret this value. What does this mean?
h. Does the sample data provide evidence to conclude that the plant is in compliance with the noise pollution code (with a = .05)?

Solution

Q6.
a)
CI = p ± Z a/2 Sqrt(p*(1-p)/n)))
x = Mean
n = Sample Size
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
CI = Confidence Interval
Mean(x)=80
Sample Size(n)=400
Sample proportion = x/n =0.2
Confidence Interval = [ 0.2 ±Z a/2 ( Sqrt ( 0.2*0.8) /400)]
= [ 0.2 - 1.96* Sqrt(0.0004) , 0.2 + 1.96* Sqrt(0.0004) ]
= [ 0.1608,0.2392]
b)
we are 95% confident that proportion rate is lies in interval
[ 0.1608,0.2392]
c)
Compute Sample Size ( n ) = n=(Z/E)^2*p*(1-p)
Z a/2 at 0.05 is = 1.96
Samle Proportion = 0.2
ME = 0.02
n = ( 1.96 / 0.02 )^2 * 0.2*0.8
= 1536.64 ~ 1537