Proportions and Percents – Assignment 4 Complete Solution

Question: #1307

Proportions and Percents – Assignment 4

One major unemployment measures identifies the proportion of the labor force that is unemployed and seeking employment. They offer the following statistics:

A sample of 625 individuals is taken.
Of the 625, 60 are seeking employment and unemployed.

Answer the questions that follow.

Develop an 80 percent confidence interval of the proportion that are unemployed and seeking employment.
State the margin of error
State the standard error
Suppose you would like your interval to have a margin of error not greater than 1 percent. What size sample should be take,
BLS seniors claim the rate could be higher than 8 percent. This likely to be the case.

Claims are made that celiac disease is present in 8 percent or more of the population. However, you celiac patients seem to represent some percent lower than 8. You sample 400 of the patients that you treated in the past two years and determine 28 had a diagnosis of celiac or were previously diagnosed for such. Answer the questions that follow with an alpha level at .05.

State the null hypothesis
State the alternative hypothesis
State the critical value
State the test statistic
State the decision
State the conclusion
State the p-value
Expose yourself to a type I or II error
Develop a 90 percent confidence interval for the data supplied.

HP Inspiron laptops occasionally fail out of the box (within 10 days). The company claims the failure rate is 1 percent. You sample 2000 Inspirons and determine 25 have failed. Answer the questions that follow.

Develop a 90 percent confidence interval for the rate of failure.
State the margin of error
Does the data support the claim made by HP. How did you determine this.
If you want your estimate to be within .5 percent, what size sample should be taken.

Jeb Bush supporters claim he has 60 percent support among registered Republicans nationally. You question this. You sample 750 republicans nationally and find his support at 400. Likely the claim is supported at a .05 level. Answer the questions that follow.

State the null hypothesis
State the alternative hypothesis
State the critical value
State the test statistic
State the decision
State the conclusion
State the p-value
Expose yourself to a type I or II error
Develop a 90 percent confidence interval for the data supplied.

Solution: #1286

Proportions and Percents – Assignment 4 Complete Solution

From the given data we can see that the sample size(n) is 625 and the sample proportion(p ̂) is 60/625 = 0.096.
Now for the 80% confidence interval the Z-multiplier is = Z_0.1=1.282
Thus the required 80% confidence interval is,
80% CI = (p ̂±Z_0.1*√((p ̂*(1-p ̂ ))/n))=(0.096±1.282*√((0.096*(1-0.096))/625))
=(0.0809,0.1111)

State the margin of error

The margin of error is = Z_0.1*√((p ̂*(1-p ̂ ))/n)=1.282*√((0.096*(1-0.096))/625) = 0.01511

State the standard error