Exam 12: Estimation: Comparing Two Populations

A study is trying to estimate the difference between the annual salaries paid to female and male employees working for the same large company. They take a random sample of 50 females and find that their average annual salary is $75 600 with a variance of $²1 250. They take a random sample of 50 males and find that their average annual salary is $78 500 with a variance of $²2 500. Find and interpret a 95% confidence interval for the difference in annual salaries for female and male employees of this large company.

Which of the following statements is correct when estimating the difference between two population proportions p₁ − p₂?

For a matched pairs experiment, find a 90% confidence interval for µ_D given that sample mean differences is 5, the standard deviation of differences is 3 and the sample sizes are 30.

We cannot estimate the difference between population means by estimating the mean difference μD, when the data are produced by a matched pairs experiment.

If two random samples of sizes 30 and 45 are selected independently from two non-normal populations with means of 53 and 57, then the mean of the sampling distribution of the sample mean difference, $\bar { X } _ { 1 } - \bar { X } _ { 2 }$ , equals -4.

The expected value of the difference of two sample means equals the difference of the corresponding population means:

Suppose that the starting salaries of male workers are normally distributed with a mean of $56 000 and a standard deviation of $12 000. The starting salaries of female workers are normally distributed with a mean of $50 000 and a standard deviation of $10 000. A random sample of 50 male workers and a random sample of 40 female workers are selected. a. What is the sampling distribution of the sample mean difference $\bar { X } _ { 1 }$ - $\bar { X } _ { 2 }$ ? Explain. b. Find the expected value and the standard error of the sample mean difference. c. What is the probability that the sample mean salary of female workers will not exceed that of the male workers?

In constructing a confidence interval estimate for the difference between two population proportions, we:

Two samples are selected at random from two independent normally distributed populations. Sample 1 has 49 observations and has a mean of 10 and a standard deviation of 5. Sample 2 has 36 observations and has a mean of 12 and a standard deviation of 3. The standard error of the sampling distribution of the sample mean difference, $\bar { X } _ { 1 } - \bar { X } _ { 2 }$ , is:

Two independent random samples are drawn from two normal populations. The sample sizes are 20 and 25, respectively. The parameters of these populations are: Population 1: \mu=505,\sigma=10 Population 2: \mu=4750,\sigma=7 Find the probability that the difference between the two sample means (X₁-bar - X₂-bar) is between 25 and 35.

When the two population variances are unequal, we cannot pool the data and produce a common estimator. We must calculate s₁² or s₂² and use them to estimate σ₁² and σ₂² respectively.

Estimate p₁ − p₂ with 99% confidence, given that n₁ = 50 and n₂ = 50 and the first sample has a proportion of 0.50 and the second sample has a proportion of 0.20

If two random samples of sizes $n _ { 1 }$ and $n _ { 2 }$ are selected independently from two non-normally distributed populations, then the sampling distribution of the sample mean difference, $\bar { X } _ { 1 } - \bar { X } _ { 2 }$ , is:

If two random samples of sizes $n _ { 1 }$ and $n _ { 2 }$ are selected independently from two populations with variances $\sigma _ { 1 } ^ { 2 }$ and $\sigma _ { 2 } ^ { 2 }$ , then the standard error of the sampling distribution of the sample mean difference, $\bar { X } _ { 1 } - \bar { X } _ { 2 }$ , equals:

In constructing a 99% confidence interval estimate for the difference between the means of two normally distributed populations, where the unknown population variances are assumed not to be equal, summary statistics computed from two independent samples are as follows: =28 x-=123 =8.5 =45 x-=105 =12.4 The lower confidence limit is:

Two independent random samples of 25 observations each are drawn from two normal populations. The parameters of these populations are: Population 1: \mu=150,\sigma=50 Population 2: \mu=130,\sigma=45 Find the probability that the mean of sample 1 will exceed the mean of sample 2.

Suppose that the starting salaries of finance graduates from university A are normally distributed with a mean of $36 750 and a standard deviation of $5320. The starting salaries of finance graduates from university B are normally distributed with a mean of $34 625 and a standard deviation of $6540. If simple random samples of 50 finance graduates are selected from each university, what is the probability that the sample mean of university A graduates will exceed that of university B graduates?

In order to draw inferences about p1 − p2, we take two independent samples − a sample of size n1 from population 1 and a sample of size n2 from population 2.

In constructing a confidence interval estimate for the difference between the means of two normally distributed populations, using two independent samples, we:

If two populations are not known to be normally distributed, the sampling distribution of the sample mean difference, $\bar { X } _ { 1 } - \bar { X } _ { 2 }$ , will be: