Question

1. Xis a normally distributed random variable with a mean of 12 and a standard deviation of 3. Calculate the probability that x equals 19.62. 2. A simple random sample of 8 employees of a corporation provided the following information 25 32 26 54 22 23 Determine the point estimate for the average age of all employees. What is the point estimate for the standard deviation of the population? Determine a point estimate for the proportion of all employees who are female. a. b. c. 3. A random sample of 36 students at a community college showed an average age of 25 years. Assume the ages of all students at the college are normally distributed with a standard deviation of 1.8 years. Compute the 98% confidence interval for the average age of all students at this college. 4. Bastien, Inc. has been manufacturing small automobiles that have averaged 50 miles per gallon of gasoline in highway driving. The company has developed a more efficient engine for its small cars and now advertises that its new small cars average more than 50 miles per gallon in highway driving. An independent testing service road-tested 36 of the automobiles. The sample showed an average of 51.5 miles per gallon. The population standard deviation is 6 miles per gallon With a 0.05 level of significance, test to determine whether or not the manufacturer's advertising campaign is legitimate. b. What is the p-value associated with the sample results?

Answer 1

Solution:-

4)

a)

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: u < 50
Alternative hypothesis: u > 50

Note that these hypotheses constitute a one-tailed test.

Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method is a one-sample z-test.

Analyze sample data. Using sample data, we compute the standard error (SE), z statistic test statistic (z).

SE = s / sqrt(n)

S.E = 1
z = (x - u) / SE

z = 1.50

z_critical = 1.645

Rejection region is z > 1.645

where s is the standard deviation of the sample, x is the sample mean, u is the hypothesized population mean, and n is the sample size.

Interpret results. Since the z-value (1.50) does not lies in the rejection region, hence we failed to reject the null hypothesis.

From the above test we do not have sufficient evidence in the favor of the claim that the manufacturer's advertising campaign is legitimate.

b)

Thus the P-value in this analysis is 0.067

Interpret results. Since the P-value (0.067) is greater than the significance level (0.05), we failed to reject the null hypothesis.