Page 173- Q 7.5
A researcher has gathered information from a random sample of 178 households. For each variable below, construct confidence intervals to estimate the population mean. Use the 90% level.
a. An average of 2.3 people resides in each household. Standard deviation is 0.35.
b. There was an average of 2.1 television sets (s _ 0.10) and 0.78 telephones (s _ 0.55) per household.
c. The households averaged 6.0 hours of television viewing per day (s _ 3.0).
A random sample of 500 residents of Shinbone, Kansas, shows that exactly 50 of the respondents had been the victims of violent crime over the past year. Estimate the proportion of victims for the population as a whole, using the 90% confi dence level. (HINT: Calculate the sample proportion P s before using Formula 7.3. Remember that proportions are equal to frequency divided by N.)
Q 8.12 page 204
A random sample of 113 convicted rapists in a state prison system completed a program designed to change their attitudes toward women, sex, and violence before being released on parole. Fifty-eight eventually became repeat sex offenders. Is this recidivism rate significantly different from the rate for all offenders (57%) in that state? Summarize your conclusions in a sentence or two. (HINT: You must use the information given in the problem to compute a sample proportion. Remember to convert the population percentage to a proportion.)
You are the head of an agency seeking funding for a program to reduce unemployment among teenage males. Nationally, the unemployment rate for this group is 18%. A random sample of 323 teenage males in your area reveals an unemployment rate of 21.7%. Is the difference significant? Can you demonstrate a need for the program? Should you use a one-tailed test in this situation? Why? Explain the result of your test of significance as you would to a funding agency.