Question

In: Statistics and Probability

(a) According to Graduating Student and Alumni Survey in 2019, among the 1230 respondents, 180 said...

(a) According to Graduating Student and Alumni Survey in 2019, among the 1230 respondents, 180 said that they expected difficulty finding a job after graduation. If we can reasonably presume that the percentage of people expected difficulty will be between 15% and 25%, obtain a sample size that will ensure a margin of error of at most 0.02 for a 90% confidence interval.

(b) A recent survey of 254 customers, selected at random from a database with 12,861 customers, found that 202 are satisfied with the service they are receiving. Find a 99% confidence interval for the percentage satisfied for all customers in the database.

(c) Independent simple random samples of 400 Hong Kong women and 380 Macao women were taken. Of Hong Kong women, 318 were found to be in the labor force; of the Macao women, 276 were found to be in the labor force. At the 5% significance level, do the data suggest that the labor-force participation rate of Hong Kong women is higher? Use the P-value approach to perform a hypothesis test.

Expert Solution

(a).

As per the provided information,

Margin of error (E) is 0.02

Confidence level __(c)__ is 0.90.

The number of respondents expected difficulty finding a job after graduation (x) is 180.

Total number of respondents (n) is 1230.

The estimated proportion is given by,

For the confidence level 0.90, the two tailed critical value obtained from the z table is +/- 1.645.

The sample size (n) can be calculated as:

Therefore, the required sample size is 845.

(b).

As per the provided information,

The total number of customers (n) is 254.

The number of customers that are satisfied with the service they are receiving (x) is 202.

The estimated proportion of customers that are satisfied with the service they are receiving is:

For the confidence level 0.99, the two tailed critical value obtained from the z table is +/- 2.576.

The 99% confidence interval for the percentage satisfied for all customers in the database can be calculated as,

Therefore, the 99% confidence interval for the percentage satisfied for all customers in the database is (0.7301, 0.8605).

orchestra answered 1 year ago

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