Question

Solve this confidence interval problem using a confidence level for that specific classmate’s problem. Make sure that you use appropriate terminology to state the problem and explain your solution.

Jennifer has 1,200 articles of clothing in her closet. She wanted to know how many skirts she owned. After cleaning out her closet, she saw that 404 out of 1,200 articles of clothing were skirts. Construct a 98% confidence interval for the proportion of clothing that turned out to be a skirt.

Please have a description of the data , the confidence level you are using, the z-critical, or t-critical value to be used in the formula, the mean or proportion that represents the data , StDev Margin of Error (MoE) - how it is calculated (formula) and the LL and UL how they were calculated (formula) OR the Stat Crunch (or Excel) Function you used along with the parameters you entered.    ALSO: The statistical statement that gives the interpretation of the results.

Answer 1

We have statement :  After cleaning out her closet, she saw that 404 out of 1,200 articles of clothing were skirts.

Here n = number of total articles . x = number of articles are skirts.

Hence the estimated proportion of articles those are skirts is :

Population proportion is normally distributed with sample size n = 1200 ( > 30). Hence the z critical value is used to find confidence interval.

The 98% confidence interval is given by,

- E < P < + E

Where,

E = margin of error.

The formula is,

c= confidence level = 0.98,

---------( using excel formula " =norm.s.inv(0.99) )

Hence the margin of error value is,

= 0.032

Hence the 98% confidence interval is,

0.34 - 0.032 < P < 0.34 + 0.032

0.308 < P < 0.372

Hence lower limit = 0.308

Upper limit = 0.372

Interpretation : We are 98% confident that population proportion of clothing that turned out to be a skirt lies between 0.308 and 0.372.