The null and alternate hypotheses are:

$$H_0:μ_1≤μ_2; H_1:μ_1>μ_2$$

A random sample of 20 items from the first population showed a mean of 100 and a standard deviation of 15. A sample of 16 items for the second population showed a mean of 94 and a standard deviation of 8. Use the .05 significant level.

The null and alternate hypotheses are:

$$H_0:μ_1=μ_2; H_1:μ_1≠μ_2$$

A random sample of 15 observations from the first population revealed a sample mean of 350 and a sample standard deviation of 12. A random sample of 17 observations from the second population revealed a sample mean of 342 and a sample standard deviation of 15. At the .10 significance level, is there a difference in the population means?

A sample of 65 observations is selected from one population with a population standard deviation of 0.75. The sample mean is 2.67. A sample of 50 observations is selected from a second population with a population standard deviation of 0.66. The sample mean is 2.59. Conduct the following test of hypothesis using the .08 significance level.a. Is this a one-tailed or a two-tailed test? b. State the decision rule. c. Compute the value of the test statistic. d. What is your decision regarding H0? e. What is the pvalue?

$$H_0:μ_1≤μ_2; H_1:μ_1>μ_2$$

The null and alternate hypotheses are:

$$\begin{aligned} & H_0: \mu_1 \leq \mu_2 \\ & H_1: \mu_1>\mu_2 \end{aligned}$$

A random sample of 20 items from the first population showed a mean of 100 and a standard deviation of 15. A sample of 16 items for the second population showed a mean of 94 and a standard deviation of 8 . Use the $.05$ significance level.

As part of a study of corporate employees, the director of human resources for PNC Inc. wants to compare the distance traveled to work by employees at its office in downtown Cincinnati with the distance for those in downtown Pittsburgh. A sample of 35 Cincinnati employees showed they travel a mean of 370 miles per month. A sample of 40 Pittsburgh employees showed they travel a mean of 380 miles per month. The population standard deviation for the Cincinnati and Pittsburgh employees are 30 and 26 miles, respectively. At the .05 significance level, is there a difference in the mean number of miles traveled per month between Cincinnati and Pittsburgh employees?

A sample of 40 observations is selected from one population with a population standard deviation of 5. The sample mean is 102. A sample of 50 observations is selected from a second population with a population standard deviation of 6. The sample mean is 99. Conduct the following test of hypothesis using the .04 significance level and Compute the value of the test statistic.

$$\begin{aligned} &H_0: \mu_1=\mu_2 \\ &H_1: \mu_1 \neq \mu_2 \end{aligned}$$

A random sample of 10 observations is selected from a normal population. The sample mean was 12 and the sample standard deviation 3. Using the .05 significance level: a. State the decision rule. b. Compute the value of the test statistic. c. What is your decision regarding the null hypothesis?

$$H_0:μ≤10; H_1:μ>10$$

A sample of 40 observations is selected from one population with a population standard deviation of 5. The sample mean is 102. A sample of 50 observations is selected from a second population with a population standard deviation of 6. The sample mean is 99. Conduct the following test of hypothesis using the .04 significance level and explain p-value?

$$\begin{aligned} &H_0: \mu_1=\mu_2 \\ &H_1: \mu_1 \neq \mu_2 \end{aligned}$$

The null and alternate hypotheses are:A sample of 100 observations from the first population indicated that X1 is 70. A sample of 150 observations from the second population revealed X2 to be 90. Use the .05 significance level to test the hypothesis. a.State the decision ruleb.Compute the pooled proportion. c. Compute the value of the test statistic. d. What is your decision regarding the null hypothesis?

$$H_0:π_1≤π_2; H_1:π_1>π_2$$

Suppose the U.S. president wants an estimate of the proportion of the population who support his current policy toward revisions in the health care system. The president wants the estimate to be within .04 of the true proportion. Assume a 95 percent level of confidence. The president’s political advisors estimated the proportion supporting the current policy to be .60. a. How large of a sample is required? b. How large of a sample would be necessary if no estimate were available for the proportion supporting current policy?

A simple random sample with $n=54$ provided a sample mean of $22.5$ and a sample standard deviation of $4.4.$

Develop a $90\%$ confidence interval for the population mean.

Suppose $6$ cans of tomato juice cost $\$ 3.20$. Find the cost of the following numbers of cans.

a. $1$ can

b. $10$ cans

c. $n$ cans

A simple random sample of $50$ items from a population with $\sigma= 6$ resulted in a sample mean of $32.$

Provide a $90\%$ confidence interval for the population mean.

A random sample of $30$ cans of Del Vino crushed tomatoes revealed a mean weight of $798.3$ grams (excluding the juice). The can-filling process for Del Vino crushed tomatoes has a known standard deviation of $3.1$ grams. Construct the $95$ percent confidence interval for the mean weight of cans, assuming that the weight of cans is a normally distributed random variable.